Holden Smith
New member
A crate of mass m = 10.1 kg is pulled up a rough incline with an initial speed of vi = 1.55 m/s. The pulling force is F = 95 N parallel to the incline, which makes an angle of ? = 19.3° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.07 m.
a) How much work is done by the gravitational force on the crate?
W = mg?h
W = mg?dsin?
W = (10.1kg)(9.8m/s²)(5.07 - 0)sin19.3 = 166J
Now, later I found out that this is wrong and they wanted me to go back to the definition of work
W = (10.1kg)(9.8m/s²)(5.07)cos(19.3 + 90) = -166J
Which kinda of make sense to me since the force of g is going down and we are going up. Then this problem came up and destroyed my intuition again
A 15.0kg box slides down an incline as shown in the diagram. If the box starts from rest at the top of the incline and has a speed of 6.0m/s at the bottom, what was the loss in mechanical energy?
?KE + ?PE + ?TE = 0
½m(v² - v?²) + mg?h + ?TE = 0
½(15kg)((6.0m/s)² - 0²) + (15kg)(9.8m/s²)(0 - 5.0m) = -?TE
270J - 735J = -?TE
?TE = 4.7 x 10²J
Notice how 735J is negative and not positive, it would have been positive if we had used the approach in the first problem where they used mg?h = mgdcos(90 + ?) would given us -735J and -(-735)J = +735J
So can anyone clarify what the first problem wants? Because I am in agreement with the second problem's approach.
a) How much work is done by the gravitational force on the crate?
W = mg?h
W = mg?dsin?
W = (10.1kg)(9.8m/s²)(5.07 - 0)sin19.3 = 166J
Now, later I found out that this is wrong and they wanted me to go back to the definition of work
W = (10.1kg)(9.8m/s²)(5.07)cos(19.3 + 90) = -166J
Which kinda of make sense to me since the force of g is going down and we are going up. Then this problem came up and destroyed my intuition again
A 15.0kg box slides down an incline as shown in the diagram. If the box starts from rest at the top of the incline and has a speed of 6.0m/s at the bottom, what was the loss in mechanical energy?
?KE + ?PE + ?TE = 0
½m(v² - v?²) + mg?h + ?TE = 0
½(15kg)((6.0m/s)² - 0²) + (15kg)(9.8m/s²)(0 - 5.0m) = -?TE
270J - 735J = -?TE
?TE = 4.7 x 10²J
Notice how 735J is negative and not positive, it would have been positive if we had used the approach in the first problem where they used mg?h = mgdcos(90 + ?) would given us -735J and -(-735)J = +735J
So can anyone clarify what the first problem wants? Because I am in agreement with the second problem's approach.