A man pulls a box ofmass 20.0 kg up a ramp inclined at an angle of 250 above the

Question

A man pulls a box ofmass 20.0 kg up a ramp inclined at an angle of 250 above the

horizontal using anapplied force of FA = 140N actingparallel to the ramp. The coefficient of

kinetic frictionbetween the box and the ramp is mk= 0.300. The box travels 3.80 m alongthe

ramp.

Calculate:

(a) The work done onthe box by the applied force FA.

ANSWER: __________________ _____

(b) The work done onthe box by the gravitational force.

ANSWER: __________________ _____

(c) The work done onthe box by the normal force.

ANSWER: __________________ _____

(d) The work done onthe box by the friction force.

ANSWER: __________________ _____

(e) The total workdone on the box.

ANSWER: __________________ _____

(f) If the speed ofthe box at the bottom of the ramp is zero, what is its speed afterit has

traveled 3.80 m alongthe ramp?

ANSWER: __________________ _____

Explanation / Answer

a. W=FA*d W=140*3.80 W=532J b. W=mgh=20*9.8*3.8sin25J c. The work done by normal force is 0 as thedisplacement and force are in different directions. d. W=fr*d W=kmgsin25*3.8 W=0.3*20*9.8*sin25*3.8J e. The resulatant force on it is F=140-fr The work done W is W=F*d W=F*3.8J f. v2-u2=2as v2-0=2*140/20*3.8 v2=53.2 v=7.294m/s Hence we get by it. The work done by normal force is 0 as thedisplacement and force are in different directions. d. W=fr*d W=kmgsin25*3.8 W=0.3*20*9.8*sin25*3.8J e. The resulatant force on it is F=140-fr The work done W is W=F*d W=F*3.8J f. v2-u2=2as v2-0=2*140/20*3.8 v2=53.2 v=7.294m/s Hence we get by it.

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