College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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1. Consider two sleds that are connected by a rope. The first sled has a mass of 50 kg, while the second sled has a mass of 60 kg. If a force of 200 N acting in the direction along the East is pulled by a rope connected to the first sled,

 A.) determine the acceleration of the sled and

B.) the tension in the rope connecting the two sleds.

 

Example is attached below. Please refer to it when answering.

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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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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