In a laboratory model of cars skidding to a stop, data are measured for four trials using two blocks. The blocks have identical masses but different coefficients of kinetic friction with a table: µ k = 0.2 and 0.8. Each block is launched with speed v i = 1 m/s and slides across the level table as the block comes to rest. This process represents the first two trials. For the next two trials, the procedure is repeated but the blocks are launched with speed v i = 2 m/s. Rank the four trials (a) through (d) according to the stopping distance from largest to smallest. If the stopping distance is the same in two cases, give them equal rank. (a) v i = 1 m/s, = 0.2 (b) v i = 1 m/s, µ k = 0.8 (c) v i = 2 m/s, = 0.2 (d) v i =2 m/s, µ k = 0.8
In a laboratory model of cars skidding to a stop, data are measured for four trials using two blocks. The blocks have identical masses but different coefficients of kinetic friction with a table: µ k = 0.2 and 0.8. Each block is launched with speed v i = 1 m/s and slides across the level table as the block comes to rest. This process represents the first two trials. For the next two trials, the procedure is repeated but the blocks are launched with speed v i = 2 m/s. Rank the four trials (a) through (d) according to the stopping distance from largest to smallest. If the stopping distance is the same in two cases, give them equal rank. (a) v i = 1 m/s, = 0.2 (b) v i = 1 m/s, µ k = 0.8 (c) v i = 2 m/s, = 0.2 (d) v i =2 m/s, µ k = 0.8
Solution Summary: The author explains how the rank of four trials based on the stopping distance is c>a=d>b.
In a laboratory model of cars skidding to a stop, data are measured for four trials using two blocks. The blocks have identical masses but different coefficients of kinetic friction with a table: µk = 0.2 and 0.8. Each block is launched with speed vi = 1 m/s and slides across the level table as the block comes to rest. This process represents the first two trials. For the next two trials, the procedure is repeated but the blocks are launched with speed vi = 2 m/s. Rank the four trials (a) through (d) according to the stopping distance from largest to smallest. If the stopping distance is the same in two cases, give them equal rank. (a) vi = 1 m/s, = 0.2 (b) vi = 1 m/s, µk = 0.8 (c) vi = 2 m/s, = 0.2 (d) vi =2 m/s, µk = 0.8
2.62 Collision. The engineer of a passenger train traveling at
25.0 m/s sights a freight train whose caboose is 200 m ahead on the
same track (Fig. P2.62). The freight train is traveling at 15.0 m/s in the
same direction as the passenger train. The engineer of the passenger
train immediately applies the brakes, causing a constant acceleration
of 0.100 m/s² in a direction opposite to the train's velocity, while the
freight train continues with constant speed. Take x = 0 at the location
of the front of the passenger train when the engineer applies the brakes.
(a) Will the cows nearby witness a collision? (b) If so, where will it take
place? (c) On a single graph, sketch the positions of the front of the pas-
senger train and the back of the freight train.
Can I get help with how to calculate total displacement? The answer is 78.3x-4.8y
2.70 Egg Drop. You are on the Figure P2.70
roof of the physics building, 46.0 m
above the ground (Fig. P2.70). Your
physics professor, who is 1.80 m tall,
is walking alongside the building at
a constant speed of 1.20 m/s. If you
wish to drop an egg on your profes-
sor's head, where should the profes-
sor be when you release the egg?
Assume that the egg is in free fall.
2.71 CALC The acceleration
of a particle is given by ax(t) =
-2.00 m/s² +(3.00 m/s³)t. (a)
Find the initial velocity Vox such that
v = 1.20 m/s
1.80 m
46.0 m
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.