When a 40-N force, parallel to the incline and directed up the incline, is applied to a crate on a frictionless incline that is 30° above the horizontal, the acceleration of the crate is 2.0 m/s^2, up the incline. The mass of the crate is ?

2 Answers
Steve J.
Apr 20, 2018

m ~= 5.8 kgm≅5.8kg

Explanation:

The net force up the incline is given by

F_"net" = m*aFnet=m⋅a

F_"net"Fnet is the sum of the 40 N force up the incline and the component of the object's weight, m*gm⋅g, down the incline.

F_"net" = 40 N - m*g*sin30 = m*2 m/s^2Fnet=40N−m⋅g⋅sin30=m⋅2ms2

Solving for m,

m*2 m/s^2 + m*9.8 m/s^2*sin30 = 40 Nm⋅2ms2+m⋅9.8ms2⋅sin30=40N

m*(2 m/s^2 + 9.8 m/s^2*sin30) = 40 Nm⋅(2ms2+9.8ms2⋅sin30)=40N

m*(6.9 m/s^2) = 40 Nm⋅(6.9ms2)=40N

m = (40 N)/(6.9 m/s^2)m=40N6.9ms2

Note: the Newton is equivalent to kg*m/s^2kg⋅ms2. (Refer to F=ma to confirm this.)

m = (40 kg*cancel(m/s^2))/(4.49 cancel(m/s^2)) = 5.8 kg

I hope this helps,
Steve

Harish Chandra Rajpoot
Jul 22, 2018

5.793\ kg

Explanation:

Given that a force F=40 \N is applied on the crate of mass m kg to cause it to move with an acceleration a=2\ \text{m/s}^2 up the plane inclined at an angle \theta=30^\circ with the horizontal.

Applying Newton's second law , the net force acting on the crate moving up the inclined plane

F_{\text{net}}=ma

F-mg\sin\theta=ma

F=m(a+g\sin\theta)

m=\frac{F}{a+g\sin\theta}

=\frac{40}{2+9.81\sin30^\circ}

=\frac{40}{6.905}

=5.793\ kg

Related questions
See all questions in Defining Force
Impact of this question
34193 views around the world
You can reuse this answer
Creative Commons License