1.) Police lieutenants, examining the scene of an accident involving two cars, m
ID: 1284354 • Letter: 1
Question
1.) Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 79 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.90. Estimate the initial speed of that car assuming a level road.
______ m/s
2.) A 28.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley (Fig. 4-57). The coefficient of static friction between the table and the block is 0.500 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.
(a) Calculate the mass of sand added to the bucket.
_______ kg
(b) Calculate the acceleration of the system.
_______ m/s2 (downward)
3.)An elevator in a tall building is allowed to reach a maximum speed of 3.5 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 3.0 m if the elevator has a mass of 1280 kg including occupants?
________ N
Explanation / Answer
1. de-acc. = u.g = 0.90 x9.8 = 8.82 m/s2
using v^2 - u^2 =2ad
0 - u^2 = 2 x -8.82 x 79
u =37.33 m/s
2.
for system just start to move. ( means acc. is justzero)
(1+m)x9.8 - 0.5 x 28 x 9.8 = (1 +28 +m)x 0
m =13 kg
b) after that kinetic friction willwork
14g - 0.320 x 28 x9.8 = (14+28)a
a = 1.18 m/s2
3. acc. needed to stop
v^2 - u^2 = 2ad
0 ^2 - 3.5^2 = 2 x a x 3
a = - 2.04 m/s 2
a is in upward. direction ( - sign means opposite to direction of velocity and velocity is downwards)
now on elevator ,
T - mg = ma
T = m ( g + a) = 1280 (9.8 = 2.04) =15155.2 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.