Overcoming friction and then some.

[anyone1979]

[SOLVED] Overcoming friction and then some.

Can anybody help me figure if I have this right?

You push a loaded wheelbarrow of mass 130 kg, exerting a horizontal force of 120 N as you do so. As the wheelbarrow moves, the frictional force acting on it is 85 N.
(a) If you begin from rest, how far do you push the wheelbarrow before reaching a moderate walking speed of 1.3 m/s?
(b) Over this distance, how much work do you do on the wheelbarrow?
(c) How much work does the frictional force do on the wheelbarrow?
(d) What is the final kinetic energy of the wheelbarrow?


a) a = 120/130 = .9m/s^2
1.3^2 = 2(.9)(x)
1.69/1.8 = .94 m

b) W = 120(.94) = 112.8J

c) W = 85(.94)(-1) = -79.9J

d) K = 1/2(130)(1.3^2) = 109.85J

 

[PhanthomJay]

Can anybody help me figure if I have this right?

You push a loaded wheelbarrow of mass 130 kg, exerting a horizontal force of 120 N as you do so. As the wheelbarrow moves, the frictional force acting on it is 85 N.
(a) If you begin from rest, how far do you push the wheelbarrow before reaching a moderate walking speed of 1.3 m/s?
(b) Over this distance, how much work do you do on the wheelbarrow?
(c) How much work does the frictional force do on the wheelbarrow?
(d) What is the final kinetic energy of the wheelbarrow?


a) a = 120/130 = .9m/s^2
1.3^2 = 2(.9)(x)
1.69/1.8 = .94 m

b) W = 120(.94) = 112.8J

c) W = 85(.94)(-1) = -79.9J

d) K = 1/2(130)(1.3^2) = 109.85J

In calculating the acceleration, you must use the NET force acting on the wheelbarrow, not just the applied force.

 

[anyone1979]

But isn't the NET force equal to the force (120) since that is the only force without a nonzero x-component?

or am I wrong?

 

[PhanthomJay]

But isn't the NET force equal to the force (120) since that is the only force without a nonzero x-component?

or am I wrong?

What about the given friction force which opposes the motion in the horizontal direction?

 

[anyone1979]

You are right, thank you.

Will this cover it? (NET force = applied force - friction force)

 

[PhanthomJay]

You are right, thank you.

Will this cover it? (NET force = applied force - friction force)

yes, that is correct, in the x direction.

 

[anyone1979]

Thank you for your help