Derek pushes a 10kg crate along a rough surface with 35N of force. He observes that it is only accelerating at a rate of 2ms2. What must the value of the force of friction be?
(Assume the only two forces acting on the object are friction and Derek).
Possible Answers:
−15N
−98N
20N
−20N
98N
Answers
Correct answer:
−15N
Explanation:
Newton's second law states that F⃗ =ma⃗ .
If Derek is pushing with 35N of force, then we should be able to solve for the acceleration of the 10kg crate.
35N=(10kg)(a⃗ )
35N10kg=a⃗
3.5ms2=a⃗
Derek observes that the crate is acceleration at a rate of 2ms2, rather than the expected 3.5ms2. An outside force is acting upon it to slow the acceleration.
The equation for the net force on the object is: F⃗ resultant=F⃗ Derek+F⃗ friction. We also know, from Newton's second law, that F⃗ resultant=ma⃗ resultant, where the resultant force and acceleration are the values actually observed.
Plug in the information we've been given so far to find the force of friction.
F⃗ resultant=F⃗ Derek+F⃗ friction
ma⃗ resultant=(35N)+F⃗ friction
(10kg)(2ms2)=(35N)+F⃗ friction
20N=35N+F⃗ friction
Subtract 35N from both sides to find the force of friction.
−15N=F⃗ friction
Friction will be negative because it acts in the direction opposite to the force of Derek.