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Anna11 [10]
3 years ago
11

suppose the same chest is at rest. you push it horizontally with force of 110N but it does not budge. What is the contact force

on the chest due to the floor
Physics
1 answer:
Radda [10]3 years ago
4 0

Answer:

110 N

Explanation:

When a force is applied on a body and body does not move, it means the body remains at rest.

In this condition, there is a contact force between the body and the floor which is called static friction.

Th static friction force is a self adjusting force and comes into play when the body is at rest.

Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.

So, the contact force between the chest and the floor is 100 N.

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Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

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V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

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s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

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Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

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2.717 m = 4.9t^2

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Hence, her feet were in the air for 0.744+0.408seconds

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V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

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3 years ago
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