Question

3) A 2 kg block is sliding at an initial speed of 10 m/s across a surface, encountering a constant friction force of 7 N. How much work is done on the block after it slides 22 cm?

Answer:  

Hint: Does the block gain or lose energy during this process? What sign does this imply for the work done on it?

4) How fast is the block moving after sliding 22 cm?

Answer:  

Hint: You can treat the block like a simple particle. What kind of energy does it have and how is this affected by the work done on it?

5) What's the total distance the block travels before coming to rest?

Correct, computer gets: 14.29 m

6) What is the average power of friction on the block over the time it takes the block to come to rest?

Answer:  


7) What was the instantaneous power of friction on the block after it slid the first 22 cm?

Answer

Answer 1

3)
Work done on the block = - Ff x distance
Where Ff is the frictional force = 7 x 0.22
= - 1.54 J

4)
Initial velocity, u = 10 m/s
Consider v as the final velocity.
Distance traveled, s = 0.22 m
Deceleration = - Ff/m = - 7/2 = - 3.5 m/s²
Using the relation, v² - u² = 2as,
v² = u² + 2as,
= 100 + 2 x (-3.5) x 0.22
= 98.46
v = SQRT[98.46]
= 9.92 m/s

5)
Initial energy = 1/2 mu² = 0.5 x 2 x 100 = 100 J
Energy spent for overcoming friction = Ff x d, where d is the stopping distance
Ff x d = 100
d = 100/7
= 14.29 m

6)
Using the relation v = u + at
t = (v - u)/a
Where v is the final velocity at rest, v = 0
u is the initial velocity, u = 10 m/s
a is the deceleration, a = - 3.5 m/s2
t = (0 - 10) / (- 3.5)
= 2.86 s
Average power = Energy/time
= 100 J / 2.86
= 35 W

7)
Time taken to complete first 0.22 m is,
t = (v - u)/a
= (9.92 - 10) / ( -3.5) = 0.022 s
Average power = Energy/time
= 1.54/0.022
= 69.73 W