Question

A 6.00 kg object is lifted vertically through a distance of 5.25 m by a light string under a tension of 80.0 N. Find: (2 marks) a. The work done by the force of tension, b. The work done by gravity, and c. The final speed of the object if it starts from rest.

Answer 1

Explanation:

It is given that,

Mas of the object, m = 6 kg

It is lifted through a distance, h = 5.25 m

Tension in the string, T = 80 N

(a) By considering the free body diagram of the object, the forces can be equated as :

$T-mg=ma$

$a=\dfrac{T-mg}{m}$

$a=\dfrac{80-6\times 9.8}{6}$

$a=3.33\ m/s^2$

Work done by tension, $W_t=F\times h$

$W_t=80\times 5.25$

$W_t=420\ J$

(b) Work done by gravity, $W_g=mgh$

$W_g=6\times 9.8\times 5.25$

$W_g=308.7\ J$

(c) Let v is the final speed of the object and u = 0

$v=\sqrt{2ah}$

$v=\sqrt{2\times 3.33\times 5.25}$

v = 5.91 m/s

Hence, this is the required solution.