Question

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50m/s . The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.400 , and the crate is pulled 5.00m .(a) How much work is done by the gravitational force on the crate?

Answer 1

167.67 Joules of work are done by the gravitational force on the crate.

Force is defined as the product of the mass of the object and acceleration. The SI unit of the force of Newton.

Gravitational force is the force applied by gravity on an object. The gravity of the earth is 9.8 m/s².

Force = Mass × Acceleration

F = m × a

Mass of the crate = 10 kg

The initial speed of the crate = 1.50 m/s

The pulling force = 100 N

Horizontal angle made by the crate = 20 °

Coefficient of kinetic friction = 0.400

The crate is pulled to a height of 5 m.

$Sin θ= \frac{height}{distance}$

$= \frac{h}{d}$

$h = dSin θ$

Work done by the gravitational force on the crate is,

Work done by gravitational force = mass × gravity × height

$W_{g} = mgh$

$W_{g} = m \times g \times d \times sinθ$

$= 10 \times 9.8 \times 5 \times sin20°$

$= 167.67 \: J$

Therefore, 167.67 Joules of work is done by the gravitational force on the crate.

To know more about force, refer to the below link:

brainly.com/question/13191643

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