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

Answer 1

Answer:

(a). The work done by the force in the first second is 0.79 J.

(b). The work done by the force in the second second is 2.40 J.

(c).  The work done by the force in the third second is 3.995 J.

(d). The instantaneous power due to the force at the end of the third second is 4.8 W.

Explanation:

Given that,

Force = 4.7 N

Mass of body = 14 kg

(a). We need to calculate the acceleration

Using formula of force

$F=ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{4.7}{14}$

$a=0.34\ m/s^2$

We need to calculate the work done by the force in the first second

Using formula of work done

$W_{1}=F\cdot s$

$W_{1}=F\times(\dfrac{1}{2}at^2)$

$W_{1}=4.7\times\dfrac{1}{2}\times0.34\times1^2$

$W_{1}=0.79\ J$

(b). We need to calculate the work done by the force in the second second

Using formula of work done

$W_{2}=F\times(\dfrac{1}{2}at^2)-W_{1}$

$W_{2}=4.7\times\dfrac{1}{2}\times0.34\times2^2-0.79$

$W_{2}=2.406\ J$

(c). We need to calculate the work done by the force in the third second.

Using formula of work done

$W_{3}=F\times(\dfrac{1}{2}at^2)-(W_{2}+W_{1})$

$W_{3}=4.7\times\dfrac{1}{2}\times0.34\times3^2-(2.406+0.79)$

$W_{3}=4.7\times\dfrac{1}{2}\times0.34\times3^2-3.1$

$W_{3}=3.995\ J$

(d). We need to calculate the instantaneous power due to the force at the end of the third second.

Using formula of power

$P=F\times v$

$P=F\times at$

Put the value into the formula

$P=4.7\times\times0.34\times3$

$P=4.8\ W$

Hence, (a). The work done by the force in the first second is 0.79 J.

(b). The work done by the force in the second second is 2.40 J.

(c).  The work done by the force in the third second is 3.995 J.

(d). The instantaneous power due to the force at the end of the third second is 4.8 W.