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Starting from rest, a 9 kg block slides 11 mdown a frictionless ramp (inclined at 30◦from the floor) to the bottom. The block thenslides an additional 20.8 m along the floorbefore coming to stop. The acceleration of gravity is 9.8 m/s^2a) Find the speed of the block at the bottom of the ramp.Answer in units of m/s.b) Find the coefficient of kinetic friction betweenblock and floor.c)Find the magnitude of the mechanical energylost due to friction.Answer in units of J.

Starting from rest, a 9 kg block slides 11 mdown a frictionless ramp (inclined at-example-1
User Elving
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1 Answer

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Part 1

Since there is no friction, there is n force opposing the downward movement of the block. The freebody diagram of the block is shown below

The force pulling the block down the ramp = mgSinθ

From the information given,

θ = 30

m = 9

g = 9.8

Thus,

Force = 9 x 9.8Sin30 = 44.1

Recall, Force = mass x acceleration

acceleration = force/mass = 44.1/9 = 4.9 m/s^2

The block accelerated to the bottom at 4.9 m/s^2

The block started from rest, thus, initial velocity = 0

Distance travelled = 11 m

We want to calculate the final velocity at the bottom. We would apply the formula,

v^2 = u^2 + 2as

where

v = final velocity

u = initial velocity

a = acceleration

s = diatance

Thus,

v^2 = 0^2 + 2 x 4.9 x 11 = 107.8

v = square root of 107.8 = 10.38

The speed of the block at the bottom of the ramp = 10.38 m/s

b) The movement of the block on the floor is opposed by friction. Fr represents frictional force. Fa is the frorce moving the block forward.

Recall, Fa = 44.1N

Frictional force = normal reaction x coefficient of friction

The block travels 20.8 m on the floor before coming to rest. Thus,

initial velocity = 10.38

final velocity = 0

We would find the acceleration by applying the formula

v^2 = u^2 - 2as

The negative sign is because the block was decelerating. Thus,

0^2 = 10.38^2 - 2 x a x 20.8

10.38^2 = 2 x a x 20.8

107.744 = 41.6

a = 107.744/41.6 = 2.59

Force with which the block moved on the floor = 9 x 2.59 = 23.31N

Recall, frictional force = 9 x 9.8 x coefficient of friction

Thus,

44.1 - 88.2 x coefficient of friction = 23.31

88.2 x coefficient of friction = 44.1 - 23.31 = 20.79

coefficient of friction = 20.79/88.2

coefficient of friction = 0.24

c) The height of the block above the ground is calculated by

height = 11 cos30 = 9.53

Potential energy = mgh

Potential energy of block at the top = 9 x 9.8 x 9.53 = 840.55 J

Kinetic energy of the block when it came to a stop = 0

Thus, Mechanical energy lost due to friction = 840.55 - 0 = 840.55 J

Starting from rest, a 9 kg block slides 11 mdown a frictionless ramp (inclined at-example-1
User Alex Brooks
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