A 3.9 kg crate is accelerated from 7.0 m/s to 9.5 m/s. What is the amount of work needed to accelerate the crate to the nearest
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Answer:
Explanation:
This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of
F - f = ma
where F is the applied Friction,
f is the frictional force acting against F,
m is the mass of the object, and
a is the acceleration of the object (NOT the velocity!)
This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation
f = μ which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation
w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:
w = 37(9.8) to 2 sig figs so
w = 360N. Now fill that in to find f:
f = (.17)(360) to 2 sig figs so
f = 61. Now for the final answer in the original equation way back up at the top:
85 - 61 = 37a and do the subtraction on the left side first:
24 = 37a and then we divide to 2 sig figs to get
a = .65 m/s/s
Since we are moving in a straight line (as opposed to on an angle) the displacement is found in
d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,
d = 2.2(3.4) to 2 sig figs so
d = 7.5 m