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dexar [7]
3 years ago
14

A 10 kg box initially at rest is pulled with a 50 N horizontal force for 4 m across a level surface. The force of friction actin

g on the box is a constant 20 N. How much work is done by the normal force pushing up on the box from the ground?
A. 10 J
B. 0 J
C. 100 J
D. 50 J
Physics
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer: 0 j

Explanation:

It is given that,

Mass of the box, m = 10 kg

Force with which the box is pulled, F = 50 N

It is moved a distance of 4 m

Force of friction acting on the box, f = 20 N

We need to find the initial kinetic energy the box have. It is clear that the box is at rest initially. As there is no motion in the box at that time. The formula for the kinetic energy of the box is given by :

As v = 0

So, the initial kinetic energy of the box is 0. Hence, this is the required solution.  

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\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{14.5}=\dfrac{1}{35}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_2}=\dfrac{1}{14.5}-\dfrac{1}{35}\\\\\dfrac{1}{R_2}=0.04039\\\\R_2=\dfrac{1}{0.04039}\\\\R_2=24.75\ \Omega

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