Answer to Question #99626 in Mechanics | Relativity for Keza vanessa

Question

We launch a 15kg block up and inclined plane forming an angle of 37 degrees with the horizontal. The block moves a distance of 3.0 on the plane before stopping.the coefficient of friction force between the block and the plane is 0.25 calculate:
a) the increase of potential energy of the block
b) the work done by the force of friction
c) the starting speed of the block
d) the speed of the block when it comes to its point of departure, descending.

Answer 1

a) The increase of potential energy of the block:

$$\Delta PE=mgh=mgs\text{ sin}\theta=\\=15\cdot9.8\cdot3\cdot\text{ sin}37^\circ=265\text{ J}.$$

b) The work done by the force of friction:

$$W_f=fs\text{ cos}\theta=\mu mgs\text{ cos}\theta=\\=0.25\cdot15\cdot9.8\cdot3\cdot\text{ cos}37^\circ=88.0\text{ J}.$$

c) The starting speed of the block can be calculated according to the conservation of energy: initially the block had only kinetic energy which then transformed to the potential energy and to the work done against friction:

$$KE=\Delta PE+W_f=353\text{ J},\\ KE=mv^2/2.$$

$$v=\sqrt{2KE/m}=\\=\sqrt{2\cdot353/15}=6.86\text{ m/s}.$$

d) The speed of the block when it comes to its point of departure, descending, can be calculated

like in c): at the highest point the speed of the block was 0 and it had its potential energy. Then this potential energy had to waste to overcome friction and increase the velocity of the block:

$$\Delta PE=KE_d+W_f,\\KE_d=\Delta PE-W_f=\\=265-88=177\text{ J},\\ \space\\ v_d=\sqrt{2KE_d/m}=\\=\sqrt{2\cdot177/15}=4.86\text{ m/s}.$$

Note: if there were no friction, that speed would be the same as in c).