Physics, amigo!!!!!!?

Question

Starting from rest, a 5 kg block slides 2.5 m down a rough 30 degree incline in 2 s.
a. What is the work done by the force of gravity
b. What is the mechanical energy lost due to friction
c. What is the work done by the normal force between the block and the incline



2) A 50 kg pole vaulter running at 10 m/s vaults over the bar. If the vaulter's horizontal component of velocity over the bar is 1 m/s and air resistance is disregarded, how high was the jump

Answer 1

(1) The initial height of the block is h = ( 2.5 m sin 30 degrees). The initial energy of the system is all potential and is equal to mgh. At the bottom of the ramp, the system energy is all kinetic, 1/2 m v^2. These energies will be different; this is the lost mechanical energy and is equal to the work done by friction.

Normal forces don't do work.

(2) We assume that the pole vaulter is able to transform all of his initial kinetic energy into work. He starts with energy 1/2 m v^2. He ends up with both potential energy (mgh) and kinetic energy (1/2 m (1 m/s)^2).

1/2 (50 kg) (10 m/s)^2 = (50 kg) (9.8 m/s^2) ( h ) + 1/2 (50 kg) (1 m/s)^2

Solve for h.

Answer 2

Q1a) Work Done= mgh= 5 x 9.81x 2.5sin30 J

b) s=ut + 0.5 a t^2
2.5= 0.5 x a x 4
a= 1.25 ms^-2
5gsin30 - f = 5a
f= 5gsin30+ 5x1.25
so mechanical energy lost= f x 2.5 J

c)WD by normal force= Fs = 0J

for qn 2, make use of projectile motion by considering the velocity in the horizontal and vertical components or u can also consider the conservation of energy in a system

Answer 3

vf = vi - gt wherein: vf = last pace vi = preliminary pace g = gravitational steady = nine.eight m/s^two t = time The ball used to be thrown vertically upwards. Its preliminary pace upon liberate is 12m/s. Once the ball is within the air, the gravity acts upon it and its pace steadily slows down because the ball is going up. When the ball reaches its maximum factor, its pace turns into 0. Using the above equation: vf = vi - gt At the ball's maximum factor, its pace is 0. zero = vi - gt vi = gt t = (vi)/g t = (12)/(nine.eight) t = one million.224 seconds ---> reply