Question

A 2.0-kg block starts with a speed of 10 m/s at the bottom of a plane inclined at 37° to the horizontal. The coefficient of sliding friction between the block and plane is µk = 0.30.(a) Use the work-energy principle to determine how far the block slides along the plane before momentarily coming to rest.(b) After stopping, the block slides back down the plane. What is its speed when it reaches the bottom? (Hint: For the round trip, only the force of friction does work on the block.)

Answer 1

Answer:

Explanation:

Kinetic energy of the block

= 1/2 x 2 x 10²

= 100 J

Let d be the length along the inclined plane upto where it can climb

Vertical height achieved

= d sin37

potential energy increased = mgH

= 2 X 9.8 X d sin37

= 11.79 d

Work done by friction on the inclined plane

= μmgcos37 x d

= .3 x 2x 9.8 x .7986d

= 4.695 d

Applying conservation of energy

4.695 d + 11.79 d = 100

d = 16.485 d = 100

d = 6.06 m

b )

In return journey , it will lose potential energy and gain kinetic energy in the face of frictional force acting on it once again.

Work done by frictional force again

= μmgcos37 x d  

= 28.45 J

Loss of potential energy

= 2 X 9.8 X d sin37

= 71.44 J

Loss of potential energy  = Gain of KE + Work done by frictional force

Gain of KE  = Loss of potential energy  - Work done by frictional force

= 71.44 - 28.45

= 42.99 J

If v be the last velocity at the bottom

1/2 mv² = 42.99

.5 x 2 x v² = 42.99

v = 6.55 m /s

Answer 2

The block cover a distance of 6.06 m when sliding and reach the bottom with a speed of 6.6 m/s when sliding back down.

INCLINED PLANE

There are 3 scenario where force or friction can be calculated on inclined plane.

1. When force applied is moving the object in opposite to the frictional force.
2. When the force is preventing the object from sliding.
3. when the friction is negligible.

Given that a  2.0-kg block starts with a speed of 10 m/s at the bottom of a plane inclined at 37° to the horizontal. The coefficient of sliding friction between the block and plane is µk = 0.30.

(a) Let us use the work-energy principle to determine how far the block slides along the plane before momentarily coming to rest.

Let us first calculate K.E of the block.

K.E = 1/2m$v^{2}$

= 1/2 x 2 x 10²

K.E = 100 J

If the length along the inclined plane is L, then the vertical height H will be

H = L sin37

The increased potential energy P.E  = mgH

P.E = 2 X 9.8 X L sin37

P.E = 11.79 L

Work done (W.D) by friction on the inclined plane = μN

Where N = μmgcos37

Then

W.D = μmgcos37 x L

Substitute all the parameters into the formula above

W.D = 0.3 x 2 x 9.8 x 0.7986L

W.D = 4.695L

According to conservation of energy,

W.D + P.E = K.E

4.695L + 11.79L = 100

16.485L = 100

L = 6.06 m

(b) Given that after stopping, the block slides back down the plane. To calculate its speed when it reaches the bottom, we need to consider it loss in potential energy and gain in kinetic energy because of frictional force acting on it.

The Work done W.D = 4.695L

W.D = 4.695 x 6.06

W.D = 28.45 J

Potential energy lost = 11.79 L

P.E = 11.79 x 6.06

P.E = 71.44 J

Gain in K.E  = Loss in P.E  - W.D by friction

= 71.44 - 28.45

= 42.99 J

But K.E = 1/2 mv² = 42.99

0.5 x 2 x v² = 42.99

v² = 42.99

v = $\sqrt{42.99}$

v = 6.55 m /s

v = 6.6 m/s approximately

Therefore, the block cover a distance of 6.06 m when sliding and reach the bottom with a speed of 6.6 m/s when sliding back down.

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