A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom o the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is 0.02. a. What is the kinetic energy of the block at the bottom of the curved surface?
b. What is the speed of the block at the bottom of the curved surface?
c. Find the stopping distance of the block?
d. Find the elapsed time of the block while it is moving on the horizontal part of the track.
e. How much work is done by the friction force on the block on the horizontal part of the track?

a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J

Explanation:

a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.

So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.

So, mgh + 0 = 0 + K'

K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m

So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J

b. Since the kinetic energy of the block K = 1/2mv²  where m = mass of block = 5 kg, v = velocity of block at bottom of curve

So, v = √(2K/m)

= √(2 × 186.2 J/5 kg)

= √(372.4 J/5 kg)

= √(74.48 J/kg)

= 8.63 m/s

c. To find the stopping distance, from work-kinetic energy principles,

work done by friction = kinetic energy change of block.

So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance

ΔK = -fd

K" - K' = - μmgd

d = -(K" - K')/μmg

Substituting the values of the variables, we have

d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)

d = -(- 186.2 J)/(0.98 kg m/s²)

d = 190 m

d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m

So, a = (v² - u²)/2d

substituting the values of the variables, we have

a = (0² - (8.63 m/s)²)/(2 × 190 m)

a = -74.4769 m²/s²/380 m

a = -0.2 m/s²

Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.

t = (v - u)/a

t =(0 m/s - 8.63 m/s)/-0.2 m/s²

t = - 8.63 m/s/-0.2 m/s²

t = 43.2 s

e. The work done by friction W = fd where

= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m

W = 0.02 × 5 kg × 9.8 m/s² × 190 m

W = 186.2 J

we need answer choices

ill do it for 100, also this isnt 80 its 40