Question

Camping equipment weighting 5000N is pulled across a frozen lake by means of a horizontal rope. There is a frictional force of 300N opposing the motion. how much work is done by the campers in pulling the equipment 1000m if its speed is increasing at the constant rate of 0.20 m/s^2? a) -2 x 10^6 J
b) 4 x 10^5 J
c) 3 x 10^5 J
d) 5.2 x 10^5 J
e) 1.2 x 10 ^6 J

Answer 1

Answer:

The work done by the campers is $4\times10^{5}\ J$

(b) is correct option.

Explanation:

Given that,

Weight = 5000 N

Frictional force = 300 n

Distance = 1000 m

Constant rate of speed = 0.20 m/s²

We need to calculate the force

Using newton's law of motion

$F-F_{\mu}=ma$

$F-300=\dfrac{5000}{10}\times0.20$

$F=\dfrac{5000}{10}\times0.20+300$

$F=400\ N$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=400\times1000$

$W=4\times10^{5}\ J$

Hence, The work done by the campers is $4\times10^{5}\ J$