Question

A frictionless cart of mass ? = 4.00 kg has a shape of a ramp with an incline angle ? = 60.0°. The ramp is pulled by a constant horizontal force ? = 15.0 N. Another frictionless cart of mass? = 2.00 kg can roll along the ramp and is attached to the top of the ramp at point ? by a spring of constant ? = 300. N/m. The spring is free to pivot without friction around point ?. Assume that the motion of the carts has stabilized, i.e. the top cart ? doesn’t oscillate with respectto ramp and the spring maintains a constant amount of stretch. Find acceleration of the carts, normal force between the carts and the amount of stretch of the spring.

Please can you do it step by step, so i can understand what is going on ?

Answer 1

Given,

mass of the inclined cart, M = 4 kg

inclination angle, = 60

mass of the small cart, m = 2 kg

Horizontal force applied, F = 15 N

Spring constant, k = 300 N/m

Let elongation in spring be x

Now,

acceleration of the combined mass, a = Force applied / total mass

                                                              = F / ( M+m) = 15/ (4+2)

                                                              = 15/6 = 2.5 m/s²

Now,

Figure above is in accelerated or non-inertial frame of reference, hence, Newton's law is not applicable here.

To convert it into inertial frame of reference, a fictious force is applied on the opposite direction such tha Net force is zero,to convert it into inertial frame of reference, where Newton's laws are applicable, as shown in figure above.

Now,

= 0

=> 0 = mgsin + masin - kx

=> kx = mgsin + macos

Put the values,

=> 300*x = 2*9.8*sin60 + 2*2.5*cos60

=> 300*x = 2*9.8*0.866 + 2*2.5*0.5 = 16.9736 + 2.5

=> x = 19.4736 / 300 = 0.0649 m

or elongation in spring is 6.49 cm

Now,

= 0

=> 0 = N + masin - mgcos

=> N = mgcos - masin

put the values,

=> N = 2*9.8*cos60 - 2*2.5*sin60

          = 2*9.8*0.5 - 2*2.5*0.866 = 5.47 N

or normal force between the carts is 5.47 N