Question

A block of mass 1.5 kg slides down an inclined plane that has an angle of 15. If the inclined plane has no friction and the block starts at a height of 3 m, how much kinetic energy does the block have when it reaches the bottom? Acceleration due to gravity is g = 9.8 m/s2.

Answer 1

`E_T = E_k_i_n + E_p_o_t = constant`

before the block slides down the energy is:

`E = mgh`
There is no kinetic energy since the block has no velocity.
When the block reaches the ground h = 0 so that the potential energy is zero and was converted into kinetic energy.

`E = (1)/(2) m v^(2)`

Since no energy is lost:

`mgh = (1)/(2) m v^(2) \\ \\ v = √(2gh)`

Answer 2

7.7 m/s

Step-by-step explanation:

At the top of the inclined plane, all the mechanical energy of the block is potential energy, given by:

`U=mgh`

where m is the mass of the block, g is the gravitational acceleration, h is the heigth.

At the bottom of the ramp, all the potential energy of the block has been converted into kinetic energy, given by:

`K=(1)/(2)mv^2`

where v is the speed of the block at the bottom of the ramp.

Since there is no friction, the energy is conserved, so the initial potential energy is equal to the final kinetic energy:

`mgh=(1)/(2)mv^2`

And we can solve the equation to find the speed of the block at the bottom of the ramp:

`gh=(v^2)/(2)\\v=√(2gh)=√(2(9.8 m/s^2)(3 m))=7.7 m/s`