Work done by friction homework help

[mandy9008]

Homework Statement

(d) Let m = 2.00 kg, x = 3.50 m, θ = 32.0°, F = 17.5 N, and µk = 0.400, and find the answers to parts (a) and (b). (Include appropriate signs.)
(a.) friction force vector f k
(b.) work done by friction
(c.) work done by vector F

Homework Equations

a. mu_k(mg-Fsinө)
b. mu_k(mg-Fsinө)x ...** wrong!
c. Fxcosө

The Attempt at a Solution

a. 4.13 N
b. -5.14 J ...*** wrong!
c. 51.94 N

 

[wwshr87]

I'm having a hard time understanding the problem, maybe you can clarify a little more.

 

[mandy9008]

I had to find equations only using symbols, then solve them. the equation for part b is wrong though

 

[wwshr87]

But, is this an incline plane problem. Where is F acting, can you sketch it?

 

[mandy9008]

Sorry, I didn't realize that I wasn't posting the entire problem


(a) A block with a mass m is pulled along a horizontal surface for a distance +x by a constant force vector F at an angle θ with respect to the horizontal. The coefficient of kinetic friction between block and table is µk. Is the force exerted by friction equal to µkmg?

Enter the expression that describes only the magnitude of the force exerted by friction. (NOTE: Use mu_k to represent either +µk or -µk, q for θ, and m, g, x, and F as appropriate.)
mu_k(mg-Fsinq)

(b) How much work is done by the friction force and by vector F ? (Don't forget the signs.) (Use mu_k for µk, q for θ, and m, g, x, and F as appropriate.)

work done by friction mu_k(mg-Fsinq)x my answer (wrong)
work done by vector F Fxcosq my answer

(d) Let m = 2.00 kg, x = 3.50 m, θ = 32.0°, F = 17.5 N, and µk = 0.400, and find the answers to parts (a) and (b). (Include appropriate signs.)

friction force vector f k 4.13 my answer
work done by friction -5.14 my answer (wrong)
work done by vector F 51.94 my answer

 

[joshmdmd]

this is pretty universal.. a is correct if x is displacement w=fd
w=mu_k(mg-fsinq)d

w is negative of course because it is acting against the movement.

 

[wwshr87]

The force exerted by friction is Ff=(µk)(mg-Fsin(theta)). Friction is doing negative work because it opposes motion. Work done by friction is -(µk)*Ff*x. Work done by the force is F*cos(angle)*x. Does this help?