A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high must a new ramp be?

To increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.

The given parameters;

Apply the principle of conservation of energy;

P.E = K.E

To increase the speed 2v at the bottom of the ramp, the height will be;

Thus, to increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.

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new height of ramp must be 4 times that of the first ramp.

Explanation:

From conservation of energy, we know that;

Potential energy at the top of ramp = kinetic energy at the bottom of ramp.

Thus;

mgh_t = ½mv²

m will cancel out to give;

gh_t = ½v²

Thus means that the height of the first ramp is directly proportional to the square of the speed.

Thus;

h_t ∝ v²

Now, for the new ramp, we are told that we want to achieve a speed of 2v at the bottom.

Thus;

h'_t ∝ (2v)²

h'_t ∝ 4v²

From earlier we saw that;

h_t ∝ v²

Thus;

New height of ramp is;

h'_t ∝ 4h_t

Thus, new height of ramp must be 4 times that of the first ramp.

continuous conducting path for current, power supply

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards