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Chapter 15: Problem 82

The 20-lb box slides on the surface for which \(\mu_{k}=0.3 .\) The box has a velocity \(v=15 \mathrm{ft} / \mathrm{s}\) when it is \(2 \mathrm{ft}\) from the plate. If it strikes the smooth plate, which has a weight of \(10 \mathrm{lb}\) and is held in position by an unstretched spring of stiffness \(k=400 \mathrm{lb} / \mathrm{ft}\), determine the maximum compression imparted to the spring. Take \(e=0.8\) between the box and the plate. Assume that the plate slides smoothly.

Short Answer

Expert verified
Apply basic principles of physics in sequence, namely work-energy, conservation of momentum during collision, and conservation of energy, to find the maximum compression of spring.

Step by step solution

01

Determine the velocity of the box when it hits the plate

The kinetic frictional force as the box slides is \( F_k = \mu_k * W = 0.3 * 20 = 6lb \). From the work-energy principle, the work done by the frictional force is equal to the change in kinetic energy of the box. In an equation, that is \( W_{12} = T_2 - T_1 \). Since friction is opposing movement, the work done by the frictional force would be negative, and we have - \( F_k * d = 1/2 * m *(v_2^2-v_1^2) \). Given that \( v_1 = 15 ft/s, m = 20/32.2 slug \) and \( d = 2 ft \), we can solve this equation to get \( v_2 \) the velocity when the box just before it hits the plate.
02

Apply the principle of conservation of momentum during the collision

The total linear momentum just before the collision equals the linear momentum just after the collision. This gives \( m1*v2 = (m1 + m2)*v3 \) where \( m1 = 20/32.2 slug \) is the mass of the box, \( m2 = 10/32.2 slug \) is the mass of the plate and \( v3 \) is the common velocity of the box and the plate just after collision. To get \( v3 \), we solve the above equation.
03

Determine the maximum compression of the spring

When box and plate move together after the collision, they will compress the spring until they come to rest momentarily. Denote \( x \) as the maximum compression of the spring, we can use the conservation of energy principle here, since all the initial kinetic energy will be converted into potential energy of the spring at maximum compression. This gives \( 1/2*(m1+m2)*v3^2 = 1/2*k*x^2. \) Given \( k = 400 lb/ft \), we solve this equation to get \( x \), the maximum compression of the spring.

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Key Concepts

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

Kinetic Friction
Kinetic friction acts between surfaces in relative motion. In this exercise, the box slides over a surface until it reaches the plate. The kinetic frictional force, denoted as \( F_k \), impacts the box's motion. It depends on the coefficient of kinetic friction \( \mu_k \) and the normal force \( W \). The formula is:
\[ F_k = \mu_k \times W \]
where \( \mu_k = 0.3 \) and \( W = 20 \text{ lb} \). This results in a kinetic friction force of 6 lb.
  • Kinetic friction opposes the relative motion.
  • It causes the box to slow as it approaches the plate.
  • Its impact is crucial for calculating how much the box will slow before reaching the plate.
Conservation of Momentum
This principle states that the total momentum of a closed system remains constant before and after an event like a collision. For the box and plate:
  • Before collision, only the box has momentum defined as \( m_1 \times v_2 \).
  • After collision, both the box and plate share a new common velocity \( v_3 \).
Utilizing the conservation of momentum:
\[ m_1 \times v_2 = (m_1 + m_2) \times v_3 \]
where \( m_1 \) and \( m_2 \) are the masses of the box and plate respectively, calculated as \( \frac{20}{32.2} \) and \( \frac{10}{32.2} \) in slugs. This equation assists in finding \( v_3 \), which influences the subsequent energy transfer to the spring.
Spring Compression
Spring compression defines how far a spring contracts under the influence of an external force. Here, once the box and plate collide and move together, they compress the attached spring. The kinetic energy resulting from their common velocity is transferred entirely to the spring as potential energy.
This relationship is expressed with:
\[ \frac{1}{2} \times (m_1 + m_2) \times v_3^2 = \frac{1}{2} \times k \times x^2 \]
where \( k \) is the spring's stiffness \( 400 \text{ lb/ft} \) and \( x \) is the maximum compression. Solving this provides the compression distance, crucial for understanding how energy transitions from motion to storage in a spring.
Work-Energy Principle
The work-energy principle describes the interaction between the work done on an object and its kinetic energy. In the context of the box sliding to the plate, the work done by friction (work-energy principle) impacts the box鈥檚 velocity before collision. This change in kinetic energy due to friction's negative work is expressed by:
\[ - F_k \times d = \frac{1}{2} \times m \times (v_2^2 - v_1^2)\]
Here,
  • \( F_k \) is the kinetic friction of 6 lb calculated earlier.
  • \( d \) is the distance (2 ft) the box slides before striking.
  • The initial velocity \( v_1 \) is \( 15 \text{ ft/s} \).
This calculation lets you determine the velocity \( v_2 \) just as the box collides with the plate, showing how friction decreases kinetic energy over distance.

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Most popular questions from this chapter

The \(200-\mathrm{kg}\) boat is powered by the fan which develops a slipstream having a diameter of \(0.75 \mathrm{~m}\). If the fan ejects air with a speed of \(14 \mathrm{~m} / \mathrm{s}\), measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest. Assume that air has a constant density of \(\rho_{w}=1.22 \mathrm{~kg} / \mathrm{m}^{3}\) and that the entering air is essentially at rest. Neglect the drag resistance of the water.

A ballistic pendulum consists of a \(4-\mathrm{kg}\) wooden block originally at rest, \(\theta=0^{\circ}\). When a 2 -g bullet strikes and becomes embedded in it, it is observed that the block swings upward to a maximum angle of \(\theta=6^{\circ} .\) Estimate the initial speed of the bullet.

A tankcar has a mass of \(20 \mathrm{Mg}\) and is freely rolling to the right with a speed of \(0.75 \mathrm{~m} / \mathrm{s}\). If it strikes the barrier, determine the horizontal impulse needed to stop the car if the spring in the bumper \(B\) has a stiffness (a) \(k \rightarrow \infty\) (bumper is rigid), and (b) \(k=15 \mathrm{kN} / \mathrm{m}\).

For a short period of time, the frictional driving force acting on the wheels of the \(2.5-\mathrm{Mg}\) van is \(F_{D}=\left(600 t^{2}\right) \mathrm{N}\) where \(t\) is in seconds. If the van has a speed of \(20 \mathrm{~km} / \mathrm{h}\) when \(t=0\), determine its speed when \(t=5 \mathrm{~s}\)

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