Asked by Anonymous
A horizontal force of 100 N pushes a 12kg block up a frictionless incline that makes an angle 25degrees with the horizontal. What is the normal force that the incline exerts on the block and what is the acceleration of the block?
I found part (a) which is the normal force. It is 149N (if I'm correct) but I don't know how to find the acceleration. Would F=ma be used in this situation?
I found part (a) which is the normal force. It is 149N (if I'm correct) but I don't know how to find the acceleration. Would F=ma be used in this situation?
Answers
Answered by
bobpursley
The normal force? mgcosTheta. Now find the force down the plane due to weight, mgSinTheta. That is down the plane, so it subtracts from 100N force
Net force= m*a
Net force= m*a
Answered by
Anthony Orso
It's intuitive
Answered by
Jean Claude
Normal force = mgcos(25) + Fsin(25)
= 12x9.8xcos(25) + 100xsin(25)
= 149 N
Acceleration is found from
F cos(25) - mg sin(25) = ma
a = [100xcos(25) - 12x9.8x sin(25)]/12 = 3.41 m/s^2
= 12x9.8xcos(25) + 100xsin(25)
= 149 N
Acceleration is found from
F cos(25) - mg sin(25) = ma
a = [100xcos(25) - 12x9.8x sin(25)]/12 = 3.41 m/s^2
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