Learning Goal: To use the principle of work and energy to determine characterist

Question

Learning Goal:

To use the principle of work and energy to determine characteristics of a system of particles, including final velocities and positions.

The two blocks shown have masses ofmA = 54 kg and mB = 65 kg . The coefficient of kinetic friction between block A and the inclined plane is ?k = 0.15 . The angle of the inclined plane is given by ? = 30 ? . Neglect the weight of the rope and pulley.

(Figure 1)

Part A - Determining the normal force acting on block A

Determine the magnitude of the normal force acting on block A, NA.

Part B - Determining the velocity of the blocks at a given position

If both blocks are released from rest, determine the velocity of block B when it has moved through a distance of s = 3.25 m .

Part C - Determining the position of the blocks at a given velocity

If both blocks are released from rest, determine how far block A has moved up the incline when the velocity of block B is (vB)2 = 6.25 m/s .

Explanation / Answer

A.) Normal force on A is given by mAgCos

Therefore, NA = mAg Cos = 54 x 9.81 x Cos30 = 458.7682974 N

B.) Friction acting on block A = NA k = 458.7682974 x 0.15 = 68.81524461 N

Energy lost to friciton after traversing a displacement of s along the plane = frictional force x s

= 68.81524461 x 3.25 = 223.649545 Joules

According to the law of conservation of energy the net energy should remain constant.

So, energy gained should be equal to energy lost.

So, Energy gained = Energy lost

i.e Potential energy gained by block A + Kinetic energy gained by blocks A and B + Energy taken for overcoming friction= Potential energy lost by block B

i.e the falling of block B is alone responsible for lifting block A and helping A to overcome the friction and the velocity gained by both the blocks

so,

mA gsSin + 0.5 (mA + mB )v2 + 223.649545 = mB g s

54 x 9.81 x 3.25 Sin 30 + 0.5 (54 + 65)v2  + 223.649545 = 65 x 9.81 x 3.25

860.8275 + 59.5 v2 + 223.649545 = 2072.3625

59.5v2 = 987.885455

v2 = 16.60311689

v = 4.074692245 m/s

C.) Since both the blocks are connected, they will have the same velocity v' = 6.25 m/s

let the mew displacement be s'

Nows using the same equation,

Potential energy gained by block A + Kinetic energy gained by blocks A and B + Energy taken for overcoming friction= Potential energy lost by block B

mA gs'Sin + 0.5 (mA + mB )v'2 + 68.81524461 x s' = mB g s'

54 x 9.81 x s' x Sin 30 + 0.5 (54 + 65) x 6.252 + 68.81524461 s' = 65 x 9.81 x s'

264.87 s' + 2324.21875 + 68.81524461 s' = 637.65 s'

2324.21875 = 303.9647554 s'

s' = 7.646342903 m

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