Term
| A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards across a friction-free surface. Determine the acceleration of the object. (Neglect air resistance.) |
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Definition
Answer: 4.95 m/s/s, right
[image]
Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 5.20 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.
a = Fnet/ m = (5.20 N, right) / (1.05 kg) =4.95 m/s/s, right |
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Term
| A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine the acceleration of the object. (Neglect air resistance.) |
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Definition
Answer: 1.82 m/s/s, right
[image]
Upon neglecting air resistance, there are four forces acting upon the object. The up and down forces balance each other. The acceleration is rightward since the rightward applied force is greater than the leftward friction force. The horizontal forces can be summed as vectors in order to determine the net force.
Fnet= ∑Fx= 5.20 N, right - 3.29 N, left = 1.91 N, right
The acceleration of the object can be computed using Newton's second law.
ax= ∑Fx/ m = (1.91 N, down) / (1.05 kg) =1.82 m/s/s, right |
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Term
| A 921-kg sports car is moving rightward with a speed of 29.0 m/s. The driver suddenly slams on the brakes and the car skids to a stop over the course of 3.20 seconds with the wheels locked. Determine the average resistive force acting upon the car. |
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Definition
Answer: 8350 N
[image]
There are three (perhaps four) forces acting upon this car. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding car. In the free-body diagram, these two forces are represented by the Ffrict arrow.
The acceleration is not given but can be calculated from the kinematic information that is given:
vi = 29.0 m/s, vf = 0 m/s, and t = 3.20 s
The acceleration of the object is the velocity change per time:
a = Delta v / t = (0 m/s - 29.0 m/s) / (3.20 s) = -9.67 m/s/s or 9.67 m/s/s, left.
This acceleration can be used to determine the net force:
Fnet = m•a = (921 kg) • (9.67 m/s/s, left) = 8350 N, left
The friction forces (surface and air) provide this net force and are equal in magnitude to this net force. |
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Term
| A 22.6-N horizontal force is applied to a 0.0710-kg hockey puck to accelerate it across the ice from an initial rest position. Ignore friction and determine the final speed (in m/s) of the puck after being pushed for a time of .0721 seconds. |
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Definition
Answer: 23.0 m/s
[image]
Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.
a = Fnet / m = (22.6 N, right) / (0.0710 kg) = 318 m/s/s, right
The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 0.0721 s) to calculate the final speed of the puck. The kinematic equation, substitution and algebra steps are shown.
vf = vi + a•t
vf = 0 m/s + (318 m/s/s)•(0.0721 s)
vf = 23.0 m/s |
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Term
| A train has a mass of 6.32x104 kg and is moving with a speed of 94.3 km/hr. The engineer applies the brakes which results in a net backward force of 2.43x105 Newtons on the train. The brakes are held for 3.40 seconds. How far (in meters) does the train travel during this time? |
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Definition
Answer: 66.8 m
[image]
There are three (perhaps four) forces acting upon this train. There is the upward force (normal force) and the downward force (gravity); these two forces balance each other since there is no vertical acceleration. The resistive force is likely a combination of friction and air resistance. These forces act leftward upon a rightward skidding train. In the free-body diagram, these two forces are represented by the Ffrict arrow. The value of this resistive force is given as 2.43x105 N. This is the net force since there are no other horizontal forces; it is the force which causes the acceleration of the train. The acceleration value can be determined using Newton's second law of motion.
a = Fnet / m = (2.43x105 N) / (6.32x104 kg) = 3.84 m/s/s, left
This acceleration value can be combined with other kinematic variables (vi = 94.3 km/hr = 26.2 m/s; t = 3.40 s) in order to determine the distance the train travels in 3.4 seconds.
d = vi • t + 0.5 •a • t2
d = (26.2 m/s) • (3.40 s) + 0.5 • (-3.84 m/s/s) • (3.40 s)2
d = 89.1 m - 22.2 m
d = 66.8 m |
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Term
| A shopper in a supermarket pushes a loaded cart with a horizontal force of 16.5 Newtons. If the cart has a mass of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest? (Neglect resistive forces.) |
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Definition
Answer: 0.419 m
[image]
Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.
a = Fnet / m = (16.5 N, right) / (33.8 kg) = 0.488 m/s/s, right
The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 1.31 s) to calculate the final speed of the cart. The kinematic equation, substitution and algebra steps are shown.
d = vi • t + 0.5 •a • t2
d = vi • t + + 0.5 • (0.488 m/s/s)•(1.31 s) 2
d = 0.419 m |
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