Question

A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time 4.50 s.(a)What is the mass of the block of ice?----- I found this to be 56.9kg and got it right.(b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ? ------ I tried using [distance=0.5at^2] but it says its wrong. how do you do this question?

Answer 1

Answer:

a) 57.0 kg b) 24.2 m

Explanation:

a) According Newton's second law, the applied force is equal to the product of the mass times the acceleration.

As the force is constant, the acceleration is constant too.

In this case, as we have as givens the distance and the time, and also we know that the block is starting form rest, we can get the acceleration as follows:

d = 1/2 * a * t² ⇒ a = 2d / t² ⇒ a= 2* 13.0 m / (4.5)² s² = 1.28 m/s²

Replacing in the Newton's 2nd Law equation:

F = m*a ⇒ m = F/a = 73.0 N / 1.28 m/s = 57.0 Kg

b) At t=4.5 s, applying the definition of acceleration, we can get the value of the velocity at that time, as follows:

v= a* t = 1.28 m/s * 4.5 s = 5.76 m/s

If the worker stops pushing at the end of the 4. 5 s, this means (neglecting friction) that from that time omwards, no net force acts on the block, so it continues moving at constant speed.

In order to get the distance moved in the next 4.20 sec, as it is moving at constant speed, we neeed just to apply the definition of velocity:

v= Δx / Δt  ⇒ Δx = v* Δt = 5.76 m/s * 4.2 m = 24.2 m

So, the total distance traveled during all the time (9.1 s) is just the sum of the 13.0 m advanced during the time when there was a constant force applied, and the last 24.2 m at constant speed, as follows:

d = 13.0 + 24.2 = 37.2 m