/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 (III) \(\mathrm{A} 2.0\) -kg blo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(III) \(\mathrm{A} 2.0\) -kg block slides along a horizontal surface with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) The block has a speed \(v=1.3 \mathrm{m} / \mathrm{s}\) when it strikes a massless spring head- on. (a) If the spring has force constant \(k=120 \mathrm{N} / \mathrm{m},\) how far is the spring compressed? (b) What minimum value of the coefficient of static friction, \(\mu_{\mathrm{S}},\) will assure that the spring remains compressed at the maximum compressed position? (c) If \(\mu_{\mathrm{s}}\) is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detach- ment occurs when the spring reaches its natural length \((x=0) :\) explain why 1

Short Answer

Expert verified
(a) Use conservation of energy, solve for compression x. (b) Use static friction balance equation and solve for \( \mu_s \). (c) Use energy conservation to find block speed on detachment.

Step by step solution

01

Determine the work done by friction

The force of friction is given by \( F_f = \mu_k \cdot m \cdot g \), where \( m = 2.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Thus, \( F_f = 0.30 \times 2.0 \times 9.8 = 5.88 \text{ N} \). This is the force opposing the block's motion.
02

Calculate the initial kinetic energy of the block

The initial kinetic energy is given by the formula \( KE = \frac{1}{2} m v^2 \). Substituting the given values, we have \( KE = \frac{1}{2} \times 2.0 \times (1.3)^2 = 1.69 \text{ J} \).
03

Set up the energy conservation equation

When the block compresses the spring completely, its kinetic energy will have been converted into spring potential energy and work done against friction. The equation is: \( KE = \frac{1}{2} k x^2 + F_f x \) Where \( x \) is the compression of the spring, \( k = 120 \text{ N/m} \), and \( F_f = 5.88 \text{ N} \).
04

Solve for spring compression distance

Substitute known values into the equation from Step 3: \( 1.69 = \frac{1}{2} \times 120 \times x^2 + 5.88 \times x \) This is a quadratic equation in the form \( ax^2 + bx + c = 0 \). Solve it to find the compression distance \( x \).
05

Find the minimum static friction coefficient

At maximum compression, the spring force \( F_s = kx \) must equal the maximum static friction force \( F_f = \mu_s mg \). Solve \( kx = \mu_s mg \) for \( \mu_s \) by substituting the previously found \( x \), force constant \( k = 120 \text{ N/m} \), and \( m = 2.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \).
06

Determine speed at detachment if \( \mu_s < \mu_s (min) \)

If \( \mu_s < \mu_s (min) \), the block will not stay compressed when the spring is at its maximum compression. When the spring decompresses and detaches the block, all potential energy converts back to kinetic energy: \( PE_{spring} = KE_{block} \). Use \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \), solve for the velocity \( v \) at detachment.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Friction
Kinetic friction is the force that opposes the movement of two surfaces sliding past each other. It acts in the direction opposite to the motion. In this physics problem involving a 2.0 kg block, the coefficient of kinetic friction is given as \( \mu_k = 0.30 \). This coefficient helps us find the frictional force exerted on the block.

To calculate the force of kinetic friction \( F_f \), use the formula:
  • \( F_f = \mu_k \times m \times g \)
  • Here, \( m = 2.0 \) kg is the mass of the block, and \( g = 9.8 \) m/s² is the acceleration due to gravity.
  • Substituting these values, we get \( F_f = 0.30 \times 2.0 \times 9.8 = 5.88 \) N.
This force plays a crucial role as it reduces the block's kinetic energy when moving over the surface, leading to the compression of the spring upon impact.
Spring Compression
Spring compression involves transforming the kinetic energy of the block into potential energy stored in the spring. When the block, moving at a speed of 1.3 m/s, hits the spring, it compresses it as it pushes against the spring force.

The spring force is determined by the spring constant \( k \), given as 120 N/m in this exercise. When the block is at its point of maximum compression, its kinetic energy \( KE \) has been converted into potential energy stored in the spring and work done against friction.
  • The total work done against friction during compression is given by \( F_f \cdot x \), where \( x \) is the compression distance.
  • The potential energy stored in the spring is expressed as \( \frac{1}{2}kx^2 \).
In this problem, the energy conservation equation setup would be \( KE = \frac{1}{2}kx^2 + F_fx \). By using this equation, you can solve for the compression distance \( x \) as the block comes to a stop, maximizing the compression of the spring.
Energy Conservation
Energy conservation is a fundamental principle that states energy cannot be created or destroyed, only transformed from one form to another. In this exercise, the block's initial kinetic energy is transformed into potential energy when compressing the spring.

Initially, the block has a kinetic energy \( KE \) equal to \( \frac{1}{2} m v^2 \), where \( m = 2.0 \) kg and \( v = 1.3 \) m/s. As the block compresses the spring, this energy is used for two things:
  • Compressing the spring, storing energy as potential energy \( \frac{1}{2}kx^2 \).
  • Doing work against the frictional force \( F_f \).
The energy conservation equation balances the block's initial kinetic energy with these two energy expenditures. If the static friction coefficient \( \mu_s \) is too low, the block will fail to remain stationary when the spring releases, leading to all the spring's potential energy reverting back to kinetic energy, causing the block to move once more. This interaction highlights the delicate balance of forces and energy transformations at play in dynamic systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A film of Jesse Owens's famous long jump (Fig. 49) in the 1936 Olympics shows that his center of mass rose 1.1 \(\mathrm{m}\) from launch point to the top of the arc. What minimum speed did he need at launch if he was traveling at 6.5 \(\mathrm{m} / \mathrm{s}\) at the top of the arc?

(1I) Draw a potential energy diagram, \(U\) vs. \(x,\) and analyze the motion of a mass \(m\) resting on a frictionless horizontal table and connected to a horizontal spring with stiffness constant \(k\) . The mass is pulled a distance to the right so that the spring is stretched a distance \(x_{0}\) initially, and then the mass is released from rest.

(III) A \(2.0-\mathrm{kg}\) block slides along a horizontal surface with a coefficient of kinetic friction \(\mu_{\mathrm{k}}=0.30 .\) The block has a speed \(v=1.3 \mathrm{~m} / \mathrm{s}\) when it strikes a massless spring head- on (as in Fig. \(8-18\) ). \((a)\) If the spring has force constant \(k=120 \mathrm{~N} / \mathrm{m},\) how far is the spring compressed? (b) What minimum value of the coefficient of static friction, \(\mu_{S}\), will assure that the spring remains compressed at the maximum compressed position? (c) If \(\mu_{\mathrm{S}}\) is less than this, what is the speed of the block when it detaches from the decompressing spring? [Hint: Detachment occurs when the spring reaches its natural length \((x=0) ;\) explain why.

(III) The potential energy of the two atoms in a diatomic (two-atom) molecule can be written $$ U(r)=-\frac{a}{r^{6}}+\frac{b}{r^{12}} $$ where \(r\) is the distance between the two atoms and \(a\) and \(b\) are positive constants. (a) At what values of \(r\) is \(U(r)\) a minimum? A maximum? (b) At what values of \(r\) is \(U(r)=0 ?\) (c) Plot \(U(r)\) as a function of \(r\) from \(r=0\) to \(r\) at a value large enough for all the features in \((a)\) and \((b)\) to show. ( \(d\) ) Describe the motion of one atom with respect to the second atom when \(E<0,\) and when \(E>0 .(e)\) Let \(F\) be the force one atom exerts on the other. For what values of \(r\) is \(F>0, F<0, F=0 ?\) (f) Determine \(F\) as a function of \(r\).

(II) A cyclist intends to cycle up a \(9.50^{\circ}\) hill whose vertical height is \(125 \mathrm{~m}\). The pedals turn in a circle of diameter \(36.0 \mathrm{~cm} .\) Assuming the mass of bicycle plus person is \(75.0 \mathrm{~kg},\) (a) calculate how much work must be done against gravity. (b) If each complete revolution of the pedals moves the bike \(5.10 \mathrm{~m}\) along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.