1. A block is given an initial speed of 10.0 m/s up a rough ramp that is inclined at 36.9° above the
horizontal. The coefficient of kinetic friction between the block and the ramp is μk = 0.2. The block
moves up the ramp, stops momentarily, and then comes back down the ramp.
a) How long does it take to go up and come back down the ramp?
b) With what speed does it get to the bottom of the ramp on its way down?
c) What is the maximum coefficient of static friction allowed between the block and the ramp in order
for the scenario outlined in the problem to occur? This part will require some thought, and you may
want to come back to it later . . .
Solution (a and b):
The force down the plane Fp = m g sinθ = m × 9.8 × sin 36.9 = 5.88 m down the plane.
The friction force is Ff = μk m g cos θ = 0.2 × m × 9.8 × cos 36.9 = 1.567 m ; direction depends on
blocks’ motion.
Going up the plane, we pick positive up, vi = + 10 m/s, ΣF = - 5.88 m + ( -1.567 m) = - 7.447 m.
Newton’s second law give ΣF = m a so - 7.447 m = m a => a = -7.447 m/s2
Then vf = vi + a t gives 0 = ( 10 + ( -7.447 tu) or tu = 1.343 s ; and the distance up the plane is
Vf2 = vi2 + 2 a Δx => 02 = 102 + ( 2 × - 7.447 × D) => D = 6.714 meters (needed below)
Going down the plane, we pick positive down, vi = 0, ΣF = 5.88 m + ( -1.567 m) = 4.313 m
Newton’s second law give ΣF = m a so 4.313 m = m a => a = 4.313 m/s2
Then vf2 = vi2 + 2 a Δx => vf2 = 02 + ( 2 × 4.313 × D) => vf = 7.61 m/s Ans. ; and
vf = vi + a t gives 7.61 = (0 + + (4.313 × td) or td =1.76 s ; the total time is 1.34 +1.76 = 3.11 s.
Ans.
Solution (c):
At top, block is instantaneously at rest, so static friction will kick in if tan θ < μs ; or
μs,max = tan 36.9 = 0.75. Ans. (Maximum since we don’t want stat. friction to kick in!)
2. A 10.0 kg block (on wheels) is pushed from behind by a 120.0 N force. A 2.00 kg block is resting
against the front (rough) face of the block. It is observed that the 2.00 kg block does not move up or
down, but rather accelerates horizontally along with the 10.0 kg block. The wheels allow you to ignore
any kinetic friction between the 10.0 kg block and the floor.
a) Draw a proper free body diagram for each block.
b) Find the minimum coefficient of static friction between the two blocks that allows this scenario to
occur.
Solution:
The 120 N provides the accelerating force for the system consisting of both blocks (12 kg). So the
acceleration of the system is 10 m/s2 right.
Considering only the 2 kg block, this means that η = 2 a = 20 N.
Again, considering only the 2 kg block, the vertical forces must add to zero: fs = mg = 19.6 N
Then, with fs < μs η , we have 19.6 < μs × 20 , or
0.98 < μs giving μs,min = 0.98 Ans.
3. A 5.00 kg cart is moving along the inside track of a vertical loop whose radius is 2.00 meters. It is
observed that at the top of the loop, the force between the cart and the track is 111 N.
a) What is the speed of the cart at the top of the loop?
b) If the cart’s speed remains constant (very difficult to do in real life), what is the force between the
cart and the loop at the bottom of the loop?
c) What minimum speed must the cart have in order to get around the loop? Assume that the cart’s
speed is constant, and this part is really independent of parts (a) and (b).
Solutions:
At the top of the loop,
mg + η = m v2/R
At the bottom of the loop,
η – mg = m v2/R
Note that η is the force between the cart and the track.
a) Top: mg + η = m v2/R : ( 5 × 9.8 ) + 111 = 5 × v2 / 2 => v = 8.00 m/s. Ans.
b) Bottom:
η – mg = m v2/R => η = mg + mv2/R = ( 5 × 9.8 ) + ( 5 × 82 / 2 ) = 209 N. Ans.
c) Top: mg + η = m v2/R : we want η > 0, so mg > m v2/R – mg => v > sq rt ( gR ) = sq rt ( 9.8 ×
2 ) = 4.43 m/s
Gives at top of loop vmin = 4.43 m/s. Ans.
4. Two blocks are connected by a massless string 1.00 meters long (you can ignore the size of the
blocks). They are on a frictionless inclined plane as shown in the figure. The lower block (M2) is 2.00
meters from the bottom of the plane, and the plane is inclined at 30° above the horizontal.
In (a) and (b), let M1 = 2.00 kg and M2 = 5.00 kg.
(a) If M1 is held in place, the system, of course, won’t move. Calculate the Tension in the string.
(b) If M1 is now released, calculate
(1) the amount of time it takes M2 to reach the bottom of the plane; and
(2) the tension in the string as the blocks slide down the plane.
(c,d) Repeat (a) and (b) if M1 = 5.00 kg and M2 = 2.00 kg. (I’m just switching the blocks.)
Solution:
For each block, the force down the plane is M g sin θ.
(a) Here, the tension in the string is equal to the force down the plane: T = 5 × g × sin 30 = 24.5 N.
Ans.
(b) Now, for M1 : T + 2 × g × sin 30 = 2 a ; and for M2 : 5 × g × sin 30 – T = 5 a
Adding these gives 7 × g × sin 30 = 7 a or a = g sin 30 = 4.9 m/s2.
Then D = ½ a t2 gives 2 = ½ × 4.9 × t2
Plugging
or t = 0.903 seconds. Ans.
a = g sin 30 into either equation gives T = 0. Ans.
(c) Here, the tension in the string is equal to the force down the plane: T = 2 × g × sin 30 = 9.8 N.
Ans.
(d) Now, for M1 : T + 5 × g × sin 30 = 5 a ; and for M2 : 2 × g × sin 30 – T = 2 a
Adding these gives 7 × g × sin 30 = 7 a or a = g sin 30 = 4.9 m/s2.
Then D = ½ a t2 gives 2 = ½ × 4.9 × t2
or t = 0.903 seconds. Ans.
Plugging a = g sin 30 into either equation gives T = 0. Ans.