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3 years ago

13

A body of mass 80kg was lifted vertically through a distance of 5.0 metres. Calculate the work done on the body. ( Acceleration

due to gravity g=10ms²)

1 answer:

sleet_krkn [62]3 years ago

8 0

Answer:

80×5×10=4000J

so therefore, work done on the body is 4000J

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A person pushes on a 57-kg refrigerator with a horizontal force of 267 N; the - sign indicates that the force points in the +x d

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(a) -267 N

Explanation: if the refrigerator is not moving, it means that the net force acting on it is zero.

We are only interested in the motion along the horizontal direction; there are two forces acting in this direction:

- The pushing force, forward, F=+267 N

- The static frictional force, backward, $F_f$

Since the net force must be zero, we have

$F+F_f =0F_f = -F = -267 N$

(b) 363.1 N

The largest pushing force that can be applied to the refrigeratore before it begins to move is equal to the magnitude of the maximum static frictional force, which is given by:

$F_f = \mu mg$

where

$\mu=0.65$ is the coefficient of static friction

m = 57 kg is the mass of the refrigerator

g = 9.8 m/s^2 is the gravitational acceleration

Substituting,

$F_f = (0.65)(57 kg)(9.8 m/s^2)=363.1 N$

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At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the

dsp73

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

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$y_b(t)=0+0.5gt^2--2$

Subtract 1 and 2 we get

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