Question

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N.

How much work does the string do on the boy if the boy stands still?

How much work does the string do on the boy if the boy walks a horizontal distance of 11 m away from the kite?

How much work does the string do on the boy if the boy walks a horizontal distance of 11 m toward the kite?

Answer 1

a) W = 0

b) W = -42.87 J

c) W = +42.87 J

Step-by-step explanation:

As we know that work done is defined as

`W = Fdcos\theta`

here F = applied force

d = displacement

Part a)

here the boy stands still to hold the string

so here displacement d = 0

so work done

W = 0 J

Part b)

Here boy run opposite to the direction of tension in string

so here work done is given as

`W = F(-d)cos\theta`

`W = 4.5(-11)(cos30)`

`W = -42.87 J`

Part c)

here boy run in the same direction in which tension is applied

so here work done is given as

`W = Fdcos\theta`

`W = 4.5(11)cos30`

`W = 42.87 J`

Answer 2

Work is a power times the separation moved toward that power
A)How much work does the string do on the kid if the kid stops?
Zero (no separation moved)
B)How much work does the string do on the kid if the kid strolls a level separation of 11 m Far from the kite?
as the kid is thought to move on a level plane, just the flat power segment can do work.
W = 4.5cos30(- 11)
W = - 42.868...
W = - 43 J
it is negative on the grounds that the power vector and the movement vector are in inverse ways.
C)How much work does the string do on the kid if the kid strolls a level separation of 11 m TOWARD the kite?
W = 43 J
Same rationale as in B yet the movement and power are both a similar way, so certain work.
The vertical segment of the power does no function as there is no movement toward that power.