Question

A block (4 kg) starts from rest and slides down a frictionless ramp #1 of height 9 m. The block then slides a horizontal distance of 1 m on a rough surface with kinetic coefficient of friction μ_k = 0.5. Next, it slides back up another frictionless ramp #2.

Find the following numerical energy values:

Initial gravitational potential energy on Ramp #1: U_1G =  J

Tries 0/3

Kinetic energy at bottom of Ramp #1 before traveling across the rough surface: K =  J

Tries 0/3

Magnitude of Heat dissipated by friction after traveling across the rough surface: f∆s =  J

Tries 0/3

Kinetic energy at bottom of Ramp #2 after traveling across the rough surface: K =  J

Tries 0/3

Final gravitational potential energy on Ramp #2 after momentarily stopping: U_2G =  J

Tries 0/3

The block then slides back down Ramp #2 and across the rough surface to Ramp #1 again. Find the following numerical energy values:

Kinetic energy at bottom of Ramp #1 after traveling across the rough surface again: K =  J

Tries 0/3

Final gravitational potential energy on Ramp #1 after momentarily stopping: U_1G =  J

Answer 1

i] Initial gravitational potential energy at ramp 1 = mgh

U1G = 4 x 9.8 x 9 = 352.8 J

ii]

Since ramp1 is frictionless, potential energy at the highest point = kinetic energy at the bottom

so, kinetic energy at the bottom of ramp1 is: K = 352.8 J

iii] Work done by friction = heat dissipated by friction

W = u_kmgL = 0.5(4)(9.8)(1) = 19.6 J

iv] Kinetic energy at the bottom of ramp2 = kinetic energy at the bottom of ramp1 - heat dissipated

=> K' = 352.8 - 19.6 = 333.2 J

v] Final gravitational potential energy at ramp2 = 333.2 J [from energy conservation]

vi] Kinetic energy at the bottom of ramp1 after traveling across the rough surface for the 2nd time is:

K'' = 333.2 - 19.6 = 313.6 J.

vii] Final gravitational potential energy on ramp1 = 313.6 J.