Question

An object with mass 2.0 kg is moving with an initial speed of u m/s on a frictionless
surface and then ramp up (also frictionless) to a height 3.75 m. If the final speed is 7 m/s,
what is its initial speed u?

Answer 1

We will apply the law of conservation of energy to calculate the initial speed. As per the law of conservation of energy the total initial energy of the system is equal to the total final energy of the system.

$K.E_{i}+P.E_{i}=K.E_{f}+P.E_{f}$

we have the kinetic energy as

$K.E=\frac{1}{2}m(v)^2$

and potential energy as

$P.E=mgh$

where m is the mass, v is the velocity and h is the height.

so as per law of conservation of energy

$\frac{1}{2}m(v_i)^2+mgh_i=\frac{1}{2}m(v_f)^2+mgh_f$

let's assume that the friction-less surface is on 0 height. so putting values in

$\frac{1}{2}(2)(u)^2+(2)(9.8)(0)=\frac{1}{2}(2)(7)^2+(2)(9.8)(3.75)$

solving it we get

$(u)^2+0=49+73.5$

$(u)^2=122.5$

taking square root on both sides we get

$u=\sqrt{122.5}$

hence initial speed is

$u=11.07\: m/s$