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

Step-by-step 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=(T-mg)/(m)`

`a=(80-6* 9.8)/(6)`

`a=3.33\ m/s^2`

Work done by tension,
`W_t=F* h`

`W_t=80* 5.25`

`W_t=420\ J`

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

`W_g=6* 9.8* 5.25`

`W_g=308.7\ J`

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

`v=√(2ah)`

`v=√(2* 3.33* 5.25)`

v = 5.91 m/s

Hence, this is the required solution.