Question

40.
Start from rest, a 2-kilogram block of wood slides a distance of two meters
down a frictionless slope, as shown.
2 kg
2 m
1 m
What is the approximate kinetic energy of the wooden block at the bottom of
the slope?
a. 400 j b. 40
c. 2003 d. 20)​

Answer 1

Answer:

We have to go about this in a few steps.

Step 1: Kinetic energy = 1/2.m.v^2

m being the mass and v the speed.

Step 2: What is its speed at the bottom of the slope? It will be accelerating frictionlessly down the slope starting from rest, so its speed after time t is

v = at, where a is the acceleration.

Step 3: Great, so what are the values for a and t? How much does it accelerate, and how long does it take to reach the bottom? That is given by the formulae:

a = g . sin U where U is the angle of incline relative to horizontal, and g is the free-fall acceleration due to gravity at the earth's surface. And

t = sqrt(2s/a) where s is the distance it has moved under acceleration a.

Putting it all together, we get

K.E. = 1/2 × m × (at)^2

= 1/2 × m x (g^2 × sin^2U) × (2s/(g×sinU))

= m × g × sinU × s

We already know the values of m (from the question) and g (constant). All we need to find are U and s.

Now we run into the first bit of ambiguity in this question. It says the plane is 20m long and 10m high. Is that 20m horizontally from one end to the other, or 20m along the slope?

A. If it is 20m along the slope, then s = 20m, and sinU = 10m/20m = 0.5. So

K.E = 2kg × 9.81m/sec^2 × 0.5 × 20m = 196.2 kg.m^2/sec^2 = 196.2 Joules.

B. On the other hand, if it is 20m horizontally, then s = 20m × cosU = 20m × 20/sqrt(10^2 + 20^2) = 40m/sqrt(5)

and sinU = 10 / sqrt(10^2 + 20^2) = 1/sqrt(5). So

K.E. = 2kg × 9.81m/sec^2 × 1/sqrt(5) × 40m/sqrt (5)

= 156.94 Joules