In the diagram above an object of mass m 250kg is sitting o

In the diagram above, an object of mass m = 2.50kg is sitting on a frictionless inclined plane which makes an angle of 60 degree with the horizontal (ground). The distance from m to the ground (along the plane) is 11.547 meters. The distance from point g to the spring (gs) is 50 meters. The object is allowed to slide

Solution

a)

Potential energy = mgh = mg*L*sin(60deg)

= 2.5*9.8*11.547*sin(60 deg)

= 245 J

b)

Conserving energy,

Potential energy at top = Kinetic energy at the bottom

So, mgh = 0.5*mv^2

So, v = sqrt(2gh)

= sqrt(2*gL*sin(60))

= sqrt(2*9.8*11.547*sin(60 deg))

= 14 m/s

c)

as no energy is lost due to friction,

so it will reach the spring with the same Kinetic energy and thus with the same velocity.

d)

Now, all the KE is converted to the spring energy

So, 0.5*mv^2 = 0.5*k*x^2

So, x = sqrt(mv^2/k)

So, x = sqrt(2.5*14^2/50000)

= 0.099 m

= 9.9 cm

e)

Now, the PE stored inititally is partly converted to work against friction and rest is convreted to KE

So, mgh = u*mgd + 0.5*mv^2

So, gL*sin(60) = u*g*d + 0.5*v^2

So, 9.8*11.547*sin(60 deg) = 0.1*9.8*50 + 0.5*v^2

So, v = 9.9 m/s

f)

Again, setting up the energy equation:

0.5*mv^2 = 0.5*k*x^2

So, x = sqrt(mv^2/k)

= sqrt(2.5*9.9^2/50000)

= 0.07 m

= 7 cm