The above Free Body Diagram represents the motion of a toy car across a floor from left to right. The weight of the .5 kg car is
about 5 N and it is being pushed with 20 N of pressure. If there is 10 N of friction across the floor, what is the Net Force acting on the car?
A
Since the car is moving to the right, the Normal Force is balancing the Weight and the Net Force is 10 N, right.
B
Friction and the Normal Force are acting opposite the Applied Force, so the Net Force is 5 N, right.
C
The Net Force is 35 N, the addition of each of the Forces acting on the car and causing it to accelerate.
D
Net Force is equal to mass times acceleration so the Net Force is .5 kg times 9.8 m/s2 or roughly 5 N.
A
Since the car is moving to the right, the Normal Force is balancing the Weight and the Net Force is 10 N, right.
B
Friction and the Normal Force are acting opposite the Applied Force, so the Net Force is 5 N, right.
C
The Net Force is 35 N, the addition of each of the Forces acting on the car and causing it to accelerate.
D
Net Force is equal to mass times acceleration so the Net Force is .5 kg times 9.8 m/s2 or roughly 5 N.
1 answer:
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Answer:
70.6 mph
Explanation:
Car A mass= 1515 lb
Car B mass=1125 lb
Speed of car B is 46 miles/h
Distance before locking, d=19.5 ft
Coefficient of kinetic friction is 0.75
Initial momentum of car B=mv where m is mass and v is velocity in ft/s
46 mph*1.46667=67.4666668 ft/s
Initial momentum of car A is given by
where
is velocity of A
Taking East as positive and west as negative then the sum of initial momentum is
The common velocity is represented as hence after collision, the final momentum is
From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence
The acceleration of two cars
From kinematic equation
hence
Substituting the value of in equation