Asked by Mike
Under the action of a constant net force, a 3.1-kg object moves a distance of 15 meters in 8.99 seconds starting from rest. What is the magnitude of this net force?
I got the answer of 0.58 Newtons, but the book says the solution is 1.2 Newtons.
To get this I found the velocity by displacement divided by time (15m/8.99s) = 1.7 m/s. Then got the acceleration by dividing velocity by time (1.7m/s / 8.99s) = 0.19m/s/s
I then plugged this acceleration into F=m*a, using the mass given in the problem to find the force. According to the book's solution however, the acceleration should be .38m/s/s, correct?
The acceleration a is
a = 2 X/t^2 = 0.371 m/s^2
The force is M a = 1.151 N
They apparently want you to round off to two significant figures, hence the 1.2.
Your mistake arises from mixing up final velocity with average velocity.
I got the answer of 0.58 Newtons, but the book says the solution is 1.2 Newtons.
To get this I found the velocity by displacement divided by time (15m/8.99s) = 1.7 m/s. Then got the acceleration by dividing velocity by time (1.7m/s / 8.99s) = 0.19m/s/s
I then plugged this acceleration into F=m*a, using the mass given in the problem to find the force. According to the book's solution however, the acceleration should be .38m/s/s, correct?
The acceleration a is
a = 2 X/t^2 = 0.371 m/s^2
The force is M a = 1.151 N
They apparently want you to round off to two significant figures, hence the 1.2.
Your mistake arises from mixing up final velocity with average velocity.
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