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A force of 4.7 N acts on a 14 kg body initially at rest. (a) Compute the work do

ID: 1673156 • Letter: A

Question

A force of 4.7 N acts on a 14 kg body initially at rest. (a) Compute the work done by the force in thefirst second.
1 J

(b) Compute the work done by the force in the second second.
2 J

(c) Compute the work done by the force in the third second.
3 J

(d) Compute the instantaneous power due to the force at the end ofthe third second.
4 W (a) Compute the work done by the force in thefirst second.
1 J

(b) Compute the work done by the force in the second second.
2 J

(c) Compute the work done by the force in the third second.
3 J

(d) Compute the instantaneous power due to the force at the end ofthe third second.
4 W

Explanation / Answer

acceleration a=F/m=4.7/14=0.336 m/s2 work done by force=gained in kinetic energy W=m*v2/2=m/2*(a*t)2 t=1 s=> W=0.789J t=2s => W=3.156J t=3s => W=7.1J work done in 1st second: 0.789 J 2nd: 3.156 -0.789=2.367 J 3rd: 7.1- 3.156=3.944J (d) the instantaneous power due to the force power=force*velocity=4.7*a*3=4.73 W t=1 s=> W=0.789J t=2s => W=3.156J t=3s => W=7.1J work done in 1st second: 0.789 J 2nd: 3.156 -0.789=2.367 J 3rd: 7.1- 3.156=3.944J (d) the instantaneous power due to the force power=force*velocity=4.7*a*3=4.73 W
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