Coefficient of friction between the block and the ramp

[myk127]

Homework Statement

A 3.50 kg block slides 3.40 m down a long 15 degrees inclines ramp. the coefficient of friction between the block and the ramp is 0.30. find the following: (a) the work done on the block by friction (b) the work done on the block by gravity (c) the work done on the block by the normal force (d) the total work done on the block. If the block has a speed of 5.0 mps at the top of the ramp, (e) find its speed after sliding 3.40 m down the ramp.

Homework Equations

The Attempt at a Solution

Wfk - work by friction
Wg - work by gravity
Wn - work by normal force
Wt - total work

(a) -0.30*(3.50*9.8*sin 15) = Wfk
(b) Wg = (3.50*9.8*cos 15) = Wg
(c) 3.5*9.8*sin 15 = Wn
(d) Wfk + Wg + Wn = Wt
(e) ?

am I correct with the solutions that I made. I feel like I made something wrong and how do you find (e)? pls tell me.

 

[LowlyPion]

Homework Statement

A 3.50 kg block slides 3.40 m down a long 15 degrees inclines ramp. the coefficient of friction between the block and the ramp is 0.30. find the following: (a) the work done on the block by friction (b) the work done on the block by gravity (c) the work done on the block by the normal force (d) the total work done on the block. If the block has a speed of 5.0 mps at the top of the ramp, (e) find its speed after sliding 3.40 m down the ramp.

The Attempt at a Solution

Wfk - work by friction
Wg - work by gravity
Wn - work by normal force
Wt - total work

(a) -0.30*(3.50*9.8*sin 15) = Wfk
(b) Wg = (3.50*9.8*cos 15) = Wg
(c) 3.5*9.8*sin 15 = Wn
(d) Wfk + Wg + Wn = Wt
(e) ?

am I correct with the solutions that I made. I feel like I made something wrong and how do you find (e)? pls tell me.

Better start with a). What is the coefficient of friction? What is it based on? Isn't it a resistance to motion based on ... what? And over what distance does it act?

 

[myk127]

oops. I forgot to multiply it by the displacement. so my solution would be:
(a) -0.30*(3.50*9.8*sin 15)(3.40) = Wfk
(b) Wg = (3.50*9.8*cos 15)(3.40) = Wg
(c) 3.5*9.8*sin 15(3.40) = Wn
(d) Wfk + Wg + Wn = Wt
(e) ?

am I right with my solutions now? how do I find (e)?

 

[LowlyPion]

oops. I forgot to multiply it by the displacement. so my solution would be:
(a) -0.30*(3.50*9.8*sin 15)(3.40) = Wfk
(b) Wg = (3.50*9.8*cos 15)(3.40) = Wg
(c) 3.5*9.8*sin 15(3.40) = Wn
(d) Wfk + Wg + Wn = Wt
(e) ?

am I right with my solutions now? how do I find (e)?

To begin with a) is incorrect. The force from friction is determined as to magnitude by the normal force from the block on the incline, even though it's effect is along the surface of the incline.

The work from gravity may be given more simply from observing its change in potential energy. How much height has it given up moving down the incline?

Work is defined by the distance over which a force acts. While there is a force normal to the block, it does not act in the direction of motion, hence it's distance is 0 and the work from the normal force is ...

d) is correct if you have the correct values fore each of the components of the total.

For the last one if you know the total work, then what else do you know from the conservation of Energy? If you have given up potential energy where has the excess over friction gone to? What kind of energy? And the formula for determining that is ...