Chapter 6: Q91P (page 147)

A block slides with constant velocity down an inclined plane that has slope angle $θ$ . The block is then projected up the same plane with an initial speed $v_{0}$ . (a) How far up the plane will it move before coming to rest? (b) After the block comes to rest, will it slide down the plane again? Give an argument to back your answer.

Short Answer

Expert verified

a) The distance at which the plane will move before coming to rest is $∆ x = \frac{v_{0}^{2}}{4 g \sin θ}$

b) when the block comes to rest, the incline is not steep enough to cause it to start slipping down the incline again.

Step by step solution

Given

Angle of slope: $θ$

Initial speed of projection: $v_{0}$

Understanding the concept

Whether the block is sliding down or up the incline, there is a frictional force in the opposite direction of the motion.The problem deals with the Newton’s second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Write the equation for net force and use the newton's second law.

Draw the free body diagram and write force equation

The free-body diagram for the first part of this problem (when the block is sliding downhill with zero acceleration) is shown next.

$m g \sin θ - f_{k} = m g \sin θ - μ_{k} F_{N} = m a_{x} = 0 m g \cos θ - F_{N} = m a_{y} = 0$

Now (for the second part of the problem, with the block projected uphill) the friction direction is reversed (see figure to the right). Newton's second law for the uphill motion leads to

$m g \sin θ + f_{k} = m g \sin θ + μ_{k} F_{N} = m a_{x} m g \cos θ - F_{N} = m a_{y} = 0$

Note that by our convention, $a_{x} > 0$ means that the acceleration is downhill, and therefore, the speed of the block will decrease as it moves up the incline.

Step 4: Calculate how far up the plane will it move before coming to rest

(a)

Using $μ_{k} = \tan (θ)$ and $F_{N} = m g \cos (θ)$ , we find the x - component of the acceleration to be,

$a_{x} = g \sin θ + \frac{μ_{k} F_{N}}{m} = g \sin θ + \frac{\tan (θ) m g \cos (θ)}{m} = 2 g \sin θ$

The distance the block travels before coming to a stop can be found by using Eq.

$v_{f}^{2} = v_{0}^{2} - 2 a_{x} ∆ x$

We get,

$∆ x = \frac{v_{0}^{2}}{2 a_{x}} = \frac{v_{0}^{2}}{4 g \sin θ}$

Figure out will it slide down the plane again after the block comes to rest

(b)

We usually expect $μ_{s} > μ_{k}$ . The “angle of repose” (the minimum angle necessary for a stationary block to start sliding downhill) is $μ_{s} = \tan (θ_{r e p o s e})$ . Therefore, we expect $θ_{r e p o s e} > θ$ found in part (a). Consequently, when the block comes to rest, the incline is not steep enough to cause it to start slipping down the incline again.