Question

A force of 100N pulls a box of 10kg to the right. The coefficient of kinetic friction is 0.1 What is the horizontal acceleration of the box?
If a vertical force of 50N acts on the box vertically upwards, then what will be the acceleration of the box?(take g=10m/s2).

Answer 1

Answer:

Explanation:  take,

                    Fnet = n x (mg - 50) = ma

                      now solve to get acc. of block

                hope it helps u ......

Answer 2

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Here's ur answer

The applied force F = 100N must balance the frictional force as well as provide the acceleration

Applied force = Frictional force + ma

Applied force = 100N

Frictional force = μN = μmg = 0.1×10×10 = 10N

Force needed for the acceleration = ma = 10a

100 = 10 + 10a

10a = 100 - 10

10a = 90

a = 9m/s²

If a vertical force of 50N acts it will increase the normal reaction, hence increasing the frictional force and hence decreases the acceleration

Frictional force = μ(mg+50) = 0.1×(100+50) = 0.1×150 = 15N

Hence,

Applied force = Frictional force + ma

100 = 15+10a

10a = 85

a = 8.5m/s²

Hope this is helpful to you

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