A 4.1-kg box is pushed along a horizontal floor by a force of magnitude 21 N at an angle θ = 35° with the horizontal. If the coefficient of kinetic friction between the block and the floor is 0.20, calculate the acceleration of the box. Be careful when calculating the normal force.
[tex]\Sigma[/tex]F = ma
Ffr= [tex]\mu[/tex]k Fn
The Attempt at a Solution
Fcos35°- Ffr = ma
21Ncos35° - [tex]\mu[/tex]k mg = ma
21Ncos35° - 0.20 (4.1 kg) (9.8 m/s2) = (4.1 kg) a
(17.2021N - 8.036N) / 4.1 kg = a
2.235 m/s2 = a
Is this correct or on the right track?