Question

A 20.0 kg box slides down a 12.0 m long incline at an angle of 30.0 degrees with the horizontal. A force of 50.0 N is applied to the box in the upward direction to try to prevent it from sliding down the incline. The applied force makes an angle of 10.0 degrees to the incline. If the incline has no friction, then the increase in the kinetic energy of the box is_________

Answer 1

Answer:

The increase in kinetic energy is 586.32 J.

Explanation:

Force acting to push the object Down the incline = component of weight acting down the incline

Let's call this force $F_D$

Force acting to push the object UP the incline = component of stopping force acting up the incline

Let's call this force $F_U$

Calculating these forces we get:

$F_D = 20*9.81 * Cos(60)$

$F_D = 98.1 N$

$F_U = 50 * Cos(10)$

$F_U = 49.24N$

The increase in the kinetic energy will be equal to the work done :

$Work = (F_D-F_U) * Distance$

$Work = (98.1 - 49.24) * 12$

$Work = 586.32 J$

Thus, the increase in kinetic energy is also 586.32 J.

Answer 2

Answer:

k.E = 538.76 Joules

Explanation:

Given that:

The mass = 20. 0 kg

Diameter (d) = 12.0 m

angle ∠ θ to the horizontal = 30°

Force (F) = 50 N

angle made by the applied force  ∠ Ф = 10°

If the incline has no friction, then the coefficient of friction (μ) = 0

Therefore the increase in the kinetic energy of the box is calculate as:

FcosФ + K.E + μmgcos(θ)d = mgsin(θ)d × sin(θ)

50cos(10) + K.E + 0×20×9.8(cos30)×12 = 20×9.8×(sin 30)×12×sin(30)

49.24 + K.E + 0 = 196×0.5×12×0.5

49.24 + K.E = 588

K.E = 588 - 49.24

k.E = 538.76 Joules

∴ the increase in the kinetic energy of the box is = 538.76 Joules