Answer:
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
