A 1.20-kg object is held 1.05 m above a relaxed, massless vertical spring with a

Question

A 1.20-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is dropped onto the spring. (a) How far does the object compress the spring? (b) Repeat part (a), but this time assume a constant air-resistance force of 0.800 N acts on the object during its motion. (c) How far does the object compress the spring if the same experiment is performed on the Moon, where g = 1.63 m/s^2 and air resistance is neglected? An inclined plane of angle theta = 20.0 degree has a spring of force constant k = 505 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.59 kg is placed on the plane at a distance d = 0.294 m from the spring. From this position, the block is projected downward toward the spring with speed V = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Explanation / Answer

4.

Using energy conservation:

KEi + PEi + Wg = KEf + PEf

Wg = m*g*sin theta*(d + x)

PE = 0.5*k*x^2

KE = 0.5*m*v^2

vi = v

vf = 0, So KEf = 0

xi = 0, so PEi = 0

xf = x

Using these values:

0.5*m*v^2 + 0 + m*g*sin theta*(d + x) = 0 + 0.5*k*x^2

rearranging this equation:

kx^2/2m - (g*sin theta)*x - (v^2/2 + g*d*sin theta) = 0

Now solving this quadratic equation:

x = [g*sin theta +/- sqrt [(g*sin theta)^2 + 4*(k/2m)*(v^2/2 + g*d*sin theta)]]/(2*k/2m)

Using given values:

x = [9.81*sin 20 deg +/- sqrt [(9.81*sin 20 deg)^2 + 2*(505/2.59)*(0.75^2/2 + 9.81*0.294*sin 20 deg)]]/(505/2.59)

since spring is compressed, so using +ve sign

x = 0.132 m

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