A m = 0.500 kg block is started at velocity and slides along a rough table ( = 0

Question

A m = 0.500 kg block is started at velocity and slides along a rough table ( = 0.300) for 1.8 m before flying horizontally off the end of the table.(Figure 1) The surface of the table is 1 m up in the air, and the block hits the ground 2.5 meters away from the base of the table.

Part A What is the velocity with which the block leaves the table? (Assume it leaves completely horizontally with no vertical

Velocity =

Part B
What is the acceleration of the block (assume the +x direction is positive) while it is sliding along the table? Acceleration =
Part C What was the initial velocity of the block?
Velocity =

Explanation / Answer

let us assume that block leaves the table with v velocity

time taken by block to reach the ground = (2/g) sec

distance travelled by block in this time = v((2/g) ) = 2.5 m (given)

so calculate v from this we have v = 2.5(g/2) =5.53 m/sec

B.

since the block was sliding on the table so the friction force acting on the block = 0.3(0.5g)

the deceleration by friction force = 0.3g

or the acceleration by friction force = -0.3g

if initial velocity was u then

v2 = u2 -2(0.3g)(1.8)

substitue value of v from part A to calculate u (initial velocity)

hope it helps :)

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