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
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?
3 answers
It's intuitive
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
0 answers