A block is being pushed forward across a frictionless surface at an angle #25°# from the horizontal. The block's mass is #15# #kg# and its acceleration is #9# #m##/##s^2# in the positive x-direction. What is the force acting on this block?

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Answer 1 · 2017-12-12T10:33:06

The force is #=149.0N#

Let the force be #=(F)N#

The angle is #theta=25^@#

The component of the force parallel to the surface is #=Fcos25^@#

The mass is #m=15kg#

The acceleration is #a=9ms^-1#

#F=ma#

Therefore,

#Fcos25^@=15*9#

#F=(15*9)/(cos25^@)=149.0N#