A block is placed at the top of a frictionless inclined plane with angle 30ø and released. The incline has a?

Question

height of 15 m and the block has mass 2 kg. How long will it take the block to reach the bottom of the incline?

Answer 1

We need to find the distance travelled by the block. A triangle with a height of 15m and angle 30 deg gives a hypothenuse of 15/Sin30 or 30m.

The only force acting to accelerate the block will be the component of gravity along the plane of motion. This equals gSin30 = 9.8*0.5 = 4.9 m/s^2

You know that it starts from rest, the acceleration and the distance travelled. Use s = u*t + 0.5*a*t^2 to work out the time taken.

Answer 2

because of the fact the incline is frictionless, power is conserved. (a) Kinetic power of the block gets switched over into its gravitational ability power => (a million/2) m (8.17)^2 = m * (9.80 one) * x sin(fifty one.5°), the place x = distance up the airplane => x = (8.17)^2 / [2 * (9.80 one) * sin(fifty one.5°) m => x = 4.35 m. (b) Time to attain the utmost distance = (vsin?) / g = 8.17 * sin(fifty one.5°) / (9.80 one) sec. = 0.sixty 5 sec. (c) on account that there is no friction, power is conserved => that is going to regain its unique velocity of 8.17 m/s on the backside of the susceptible airplane.