/*! 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 57 A block is shoved up a \(22^{\ci... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 57

A block is shoved up a \(22^{\circ}\) slope with an initial speed of \(1.4 \mathrm{m} / \mathrm{s}\) The coefficient of kinetic friction is \(0.70 .\) (a) How far up the slope will the block get? (b) Once stopped, will it slide back down?

Short Answer

Expert verified
Stated calculations will give the distance up the slope the block will get. Without information regarding the coefficient of static friction, it can't be definitively determined if the block will slide back down.

Step by step solution

01

Determine force due to gravity

Firstly, the force due to gravity (Fg) acting on the block can be calculated using the equation \(Fg = m \cdot g \cdot \sin(\theta)\) where \(m\) is the mass of the block, \(g = 9.8 m/s^2\) is the acceleration due to gravity and \(\theta = 22^\circ\) is the angle of the slope. The mass of the block is not given but will soon cancel out.
02

Determine force due to kinetic friction

Next, the force due to friction (Ff) can be calculated using the equation \(Ff = \mu_k \cdot m \cdot g \cdot \cos(\theta)\) where \(\mu_k = 0.70\) is the coefficient of kinetic friction, \(m\) is the block's mass, \(g = 9.8 m/s^2\) is the acceleration due to gravity and \(\theta = 22^\circ\) is the angle of the slope. The mass of the block is not given but will soon cancel out.
03

Determine net force on the block

The net force acting on the block can be determined by subtracting the force of friction from the gravitational force \(F_{net} = Fg - Ff\). Here the mass of the block finally cancels out.
04

Apply the kinematic equation

The distance the block travels before it stops can now be found by applying the kinematic equation \(v_f^2 = v_i^2 - 2 \cdot a \cdot d\). Where \(v_f = 0 m/s\) is the final velocity (because the block stops), \(v_i = 1.4 m/s\) is the initial velocity, \(a = F_{net} / m\) is the deceleration due to the net force (which can also be seen as acceleration because of the direction) and \(d\) is the distance we want to find.
05

Will the block slide back down?

The block will slide back down the slope if the static friction force is less than the component of the weight down the slope. Since we do not have information about static friction in the problem, we cannot answer this question definitively.

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