Question

A 10.0 kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle. If the coefficient of kinetic friction is 0.30. Calculate the acceleration.

Answer 1

1.66 m/s²

Step-by-step explanation:

mass of box, m = 10 kg

Force, F = 40 N

Angle of inclination, θ = 30°

coefficient of friction, μ = 0.3

Let N be the normal reaction,

equilibrium of force in y axis

N + F Sinθ = mg

N = 10 x 9.8 - 40 Sin 30

N = 98 - 20 = 78 N

Friction force, f = μN = 0.3 x 78 = 23.4 N

Now, use Newton's second law in X axis

F - f = ma

where, a be the acceleration

40 - 23.4 = 10 x a

a = 1.66 m/s²

Thus, the acceleration is 1.66 m/s².

Answer 2

Step-by-step explanation:

Given

mass of box
`m=10\ kg`

Force applied
`F=40 \N`

inclination of force
`\theta =30^(\circ)` (w.r.t horizontal)

coefficient of kinetic friction
`\mu _k=0.3`

Net Normal reaction on the box is

`N=mg-F\sin \theta`

Friction force
`f_r=\mu _kN`

`f_r=0.3(10* 9.8-40* \sin 30)`

`f_r=23.4\ N`

force which is moving the box

`F\cos \theta`

Net force on the block

`F\cos \theta -f_r=ma`

`40* \cos 30-23.4=10* a`

`a=1.12\ m/s^2`