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
Apr 20, 2018

Answer:

#m ~= 5.8 kg#

Explanation:

The net force up the incline is given by

#F_"net" = m*a#

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

#F_"net" = 40 N - m*g*sin30 = m*2 m/s^2#

Solving for m,

# m*2 m/s^2 + m*9.8 m/s^2*sin30 = 40 N#

#m*(2 m/s^2 + 9.8 m/s^2*sin30) = 40 N#

#m*(6.9 m/s^2) = 40 N#

#m = (40 N)/(6.9 m/s^2)#

Note: the Newton is equivalent to #kg*m/s^2#. (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

Answer:

#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#