# Work done by string, friction, pulling mass up an inclined plane

1. Mar 15, 2011

### revere21

1. The problem statement, all variables and given/known data
A block of mass 1.60 kg is pulled up a rough 30 degree incline at constant speed by a string parallel to the surface. The friction force on the block is 7.00 N. The incline is 4.00 m long and 2.00 m high.

(1) How much gravitational potential energy did the block gain?
(2) How much work was done by friction?
(3) How much work was done by the string?

2. Relevant equations
PEG = m*g*y
W = F(parallel)*d

3. The attempt at a solution
(1) [using equation 1] The gravitational potential energy gained was just (1.6 kg)(9.8 m/s2)(2 m) = 31.4 kg*m2/s2, which is 31.4 J.

(2) [using equation 2] The work done by friction was (7 N)(4 m) = 28 J. BUT, the answer key says -28 J, so my best reasoning for why is that, the friction force of 7 N is opposite the direction of motion, which makes this value negative. Can someone please tell me if my line of thinking is correct, and add an explanation that will allow me to (in the future) understand when the value should be negative and when it should not?

(3) For the amount of work done by the string, I really have no idea. It would need to be greater than the 28 J of work done by friction in order to pull the object up the incline (forgive me if this is a gross oversimplification), and it would also be combatting gravity. And surely the mass must be figured in as well. And, I believe, the distance traveled. Can anyone help me in setting up an equation that would correctly integrate everything that must be included and lead me in the right direction?

2. Mar 15, 2011

### revere21

(1) Why is the work done on the block by the normal force zero?
I believe this would be because the normal force of the block is equal to the weight, but I am not sure.

(2) Imagine the string is now released. The block starts from rest and slides down the rough plane. At the bottom, its kinetic energy is:
I put that it is equal to its potential energy at the top. My reasoning is due to conservation of energy.

Thank you all again.

3. Mar 17, 2011

### Tong32

(2) The work done by friction on the block is intrinsically negative.
(3) W = F x d W = (mgsinθ + Ff) x d (1.6kg(sin30) + 7.0N) x 4.0 m = 59.4
Since the block is moving at a constant speed we assume the acceleration is zero. This then allows us to know that the force normal is equal to the opposing forces (friction and the force of gravity in the x direction).

4. Mar 17, 2011

### Tong32

(1) Because it is perpendicular to the direction of motion
(2) The kinetic energy is less than its potential energy at the top because some of the energy is transferred into heat due to friction.