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- Homework Statement
- Students are running an experiment to determine the mass of a block without using a scale. The students set up a block on a horizontal surface of negligible friction. They then attach a force sensor to the block that is at rest. The sensor is then used to exert a force on the block, moving it along a straight line. The force exerted as a function of time is shown in the top graph below. At time t = 12 s, the block is moving at 1.5 m/s.

(a)

i. Explain how the students could use the graph to estimate the impulse exerted on the block.

ii. Calculate the estimated impulse exerted on the block for the time interval 0 to 12 seconds.

(b) Calculate the mass of the block.

When t = 12s, the force sensor loses contact with the block, which continues to move along the horizontal surface. When t = 20s, the block encounters a rough area and comes to rest in 25 cm.

(c) On the (bottom) graph, sketch a graph of the speed of the block as a function of time.

d) Calculate the coefficient of kinetic friction between the block and the rough surface.

- Relevant Equations
- p = mv

J = Ft

F = ma

(bottom graph relates only to c)

(a)

i. The students can calculate the area under the graph to find the impulse exerted on the block. This is because the area under a force vs. time graph is the change in momentum or the impulse.

ii. Knowing that the graph is linear and begins at around 3 N when t = 0s, the function F(t) = 3 + t/4 can be used. When taking the integral from 0 to 12 of this function, an impulse of 54 kg * m/s is given.

(b)

I = mv

54 = (1.5)m

m = 34 kg

(c) (unsure of this one)

From 0-20s, there is a linearly increasing force on the block, meaning that the acceleration of the block is increasing linearly, so the speed is increasing exponentially. However, after 20 s, the block goes over a rough surface, causing it to slow down and eventually stop over a much shorter time interval.

(d) (unsure of this one as well)

Assuming that the function described in aii still holds true after 12 s but before 20 s, then the force at 20 s can be found with F(20). This means that at 20s, there is a force of 8 N being applied to the box. The box slows down to a stop after 20 s, so:

Fnet = Ff (friction force)

8 = mgμ

8 = (34)(9.8)μ

μ = 0.024 ???

If anyone could help with these, that would be greatly appreciated!