A m = 0.500 kg block is started at velocity vo and slides along a rough table (k

Question

A m = 0.500 kg block is started at velocity vo and slides along a rough table (k = 0.300) for 1.1 m before flying horizontally off the end of the table.(Figure 1) The surface of the table is 0.9 m up in the air, and the block hits the ground 3.1 meters away from the base of the table.

Part AWhat is the velocity with which the block leaves the table? (Assume it leaves completely horizontally with no vertical component to the velocity)

Part B

What is the acceleration of the block (assume the +x direction is positive) while it is sliding along the table?

Part C

What was the initial velocity v0 of the block?

Explanation / Answer

Part A)

We need the time to fall

d = vot + .5at2

.9 = 0 + .5(9.8)(t2)

t = .4286 sec

Then use that time to find the x velocity

d = vt

3.1 = v(.4286)

v = 7.23 m/s

Part B)

The work done by friction = umgd

W = (.3)(.5)(9.8)(1.1) = 1.617 J

That is the change in KE of the block

1.617 = .5m(v2 - v2)

1.617 = .5(.5)(v2 - 7.232)

v = 7.66 m/s

Then vf2 = vo2 + 2ad

7.232 = 7.662 + 2(a)(1.1)

a = -2.91 m/s2 (Negative since its slowing down)

Part C)

The initial velocity was found in part B.

That is 7.66 m/s

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