A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees and the coefficient of dynamic friction is 0.72, what is the speed
with which the box is moving with, assuming it takes 4seconds to reach the top of the incline?

19.6 m/s

Explanation:

The parameters given are:

Mass M = 20 Kg

Force F = 285 N

Angle Ø = 30 degree

Time t = 4 seconds

Coefficient of friction = 0.72

At the plane, the weight of the box will be mgsinØ

Resolving forces at the plane, we will have:

MgsinØ + Fr = F

Where Fr = frictional force.

Fr = F - mgsinØ

Substitute all the parameters into the formula

Fr = 285 - 20 × 9.8 sin30

Fr = 285 - 98

Fr = 187 N

But for the box moving toward the top of the plane,

F - Fr = ma

Where a = V/t

Substitute all the parameters involved into the formula

285 - 187 = 20 ( V/4)

98 = 5V

V = 98/5

V = 19.6 m/s

Therefore, the speed with which the box is moving is 19.6 m/s

multiple

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.