To push a 21 kg crate up a frictionless incline, angled at 40° to the horizontal, a worker exerts a force of 224.9 N, parallel t
1 answer:
Answer:
Work done, W = 337.35 Joules
Explanation:
It is given that,
Mass of the crate, m = 21 kg
Force exerted by the worker on the crate, F = 224.9 N
The crate is inclined at an angle of 40 degrees.
Distance covered by the slide, d = 1.5 m
Let W is the work is done on the crate by the worker's applied force. it is equal to the product of force and distance. It is given by :
W = 337.35 Joules
So, the work is done on the crate by the worker's applied force is 337.35 Joules. Hence, this is the required solution.
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, 
, so we can write:
(1)
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find