Answer to Question #129389 in Physics for allan

Question

A 8 kg block was placed on an inclined plane (35 degrees with respect to the horizontal). A 150 N horizontal force is applied on the block. The coefficients of static and kinetic friction are 0.3 and 0.2, respectively. The block was originally 1 meter above the ground and was moved 1.5 meters vertically from its initial point in a period of 2 seconds. a) Determine the amount of work done. b) What is its initial potential energy? c) What is its final potential energy? d) How much is the change in potential energy? e) How much power was used to move the block 1.5 meters vertically?

Answer 1

b) The initial potential energy is:

$U_i = mgh_i$

where $m = 8kg$ is the mass of the block and $h_i = 1m$ is its initial height. Thus:

$U_i = mgh_i = 8\cdot 9.81\cdot 1 =78.48J$

c) The final potential energy is:

$U_f = mgh_f$

where $m = 8kg$ is the mass of the block and $h_f = 1.5m$ is its initial height. Thus:

$U_f = mgh_f = 8\cdot 9.81\cdot 1.5 =117.72J$

d) The change in potential energy is:

$\Delta U = U_f - U_i = 117.72 - 78.48 = 39.24J$

a) The work against the gravity force is equal to the change in potential energy. Thus:

$A = \Delta U = 39.24J$

e) The power, by definition, is:

$P = \dfrac{A}{t} = \dfrac{39.24J}{2s} = 19.62W$

Answer. (a) 39.24 J, (b) 78.48 J, (c) 117.72 J, (d) 39.24 J, (e) 19.62 W.