204k views
4 votes
A block of mass 3.5 kg, which has an initial velocity of 6.3 m/s at time t = 0, slides on a horizontal surface.

1) Calculate the work that must be done on the block to bring it to rest. Answer in units of J.
2) If a constant friction force of 7.7 Newtons is exerted on the block by the surface, what is the acceleration? Answer in units of m/s 2 .
3) Determine the distance that the block slides as it comes to rest. Answer in units of m.

User Josh Bedo
by
8.3k points

1 Answer

4 votes

Final answer:

1) The work done on the block to bring it to rest is 69.95 J. 2) The acceleration of the block is 2.2 m/s^2. 3) The distance that the block slides as it comes to rest is 14.3 m.

Step-by-step explanation:

1) To calculate the work done on the block to bring it to rest, we need to use the work-energy principle. The work done is equal to the change in kinetic energy of the block. Since the block comes to rest, its final kinetic energy is zero. Therefore, the work done on the block is equal to the initial kinetic energy of the block. The formula for kinetic energy is KE = 1/2 * mass * velocity^2. Plugging in the values, we get KE = 1/2 * 3.5 kg * (6.3 m/s)^2 = 69.95 J.

2) The acceleration of the block can be found using Newton's second law, F = ma. The friction force is given as 7.7 N. The net force acting on the block is the friction force. So, 7.7 N = 3.5 kg * a. Solving for a, we get a = 2.2 m/s^2.

3) The distance that the block slides as it comes to rest can be found using the kinematic equation, v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (6.3 m/s), a is the acceleration (-2.2 m/s^2), and s is the distance traveled. Plugging in the values, we get 0 = (6.3 m/s)^2 + 2 * (-2.2 m/s^2) * s. Solving for s, we get s = 14.3 m.

User VelLes
by
7.3k points