The coefficient of sliding friction between rubber tires and the wet pavement is 0.50. The brakes are applied to a 2500-kg jeepney traveling 20 m/s (East), and the jeepney skids to stop. A.) What is the size and direction of the force of friction that the road exerts on the jeepney? B.) What would be the size and direction of the acceleration of the jeepney?
The coefficient of sliding friction between rubber tires and the wet pavement is 0.50. The brakes are applied to a 2500-kg jeepney traveling 20 m/s (East), and the jeepney skids to stop. A.) What is the size and direction of the force of friction that the road exerts on the jeepney? B.) What would be the size and direction of the acceleration of the jeepney?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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1. The coefficient of sliding friction between rubber tires and the wet pavement is 0.50. The brakes are applied to a 2500-kg jeepney traveling 20 m/s (East), and the jeepney skids to stop.
A.) What is the size and direction of the
B.) What would be the size and direction of the acceleration of the jeepney?
Example is attached below. Please refer to it when answering.
Transcribed Image Text:2. Jemuel (30kg), starting from rest, gives his brother John Paul (30kg) a ride
on a sled by exerting a force of 300 N (East) for 8.0 seconds while a frictional
force of 100 N is acting on the opposite direction. Determine the following:
a. acceleration of the sled together with Peter and his sister.
b. the final velocity of the sled with Jemuel and John Paul when it
reached 8.0s.
c. distance traveled by the sled with Jemuel and John Paul in 8.0s.
I.
Given: V1 = 0
a. a = ?
At = 8.0 s
a = 0
b. Vr = ?
c. d = ?
Fa = 300 N East
Ff = 200 N West
m(combined) = 60kg
II.
Free – body diagram:
N = 588 N
Ff = 100 N
Fa = 300 N
W = 588 N
III.
Solution:
b. Vi = V1 + at
= 0 + (1.67 m/s2) (8.0 s)
= 13.36 m/s
c. d = Vịt + ½ at²
= 0 + ½ (1.67 m/s2) (8.0 s)2
= ½ (1.67 m/s2) 64 s2)
a. Fnet = Fa+Ff
= 300N + (-200N)
= 100N
Fnet=ma
a = Fnet/m
= 100N/60kg
= 1.67m/s?
= 53.44 m
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