(a) The block has a weight of (27.0 kg) <em>g</em>, and the normal force of the surface pushing upward on the block has the same magnitude, so that static friction exerts a maximum force of
<em>µ</em> (27.0 kg) <em>g</em> = 70.0 N
where <em>µ</em> is the coefficient of <u>static</u> friction. Solving for <em>µ</em> gives
<em>µ</em> = (70.0 N) / ((27.0 kg) <em>g</em>) ≈ 0.265
(b) As it's moving, the block still has the same weight and thus feels the same normal force, (27.0 kg) <em>g</em>. In order to move at a constant speed, kinetic friction must exert the same force as the push, so
<em>µ</em> (27.0 kg) <em>g</em> = 64.0 N
where <em>µ</em> is now the coefficient of <u>kinetic</u> friction. Solve for <em>µ</em> :
<em>µ</em> = (64.0 N) / ((27.0 kg) <em>g</em>) ≈ 0.242
Short answer: v = 10 m/s
Comment
This is just an application of KE = 1/2 m v^2
Givens
KE = 100 Joules
m = 2 kg
v = ???
Formula
KE = 1/2 m v^2
Substitution and Solve
100 = 1/2 * 2 * v^2
100 = v^2 Take the square root of both sides.
sqrt(100) = sqrt(v^2)
v = 10 m/s
Comment
There is no KE unless v > 0.
You may have PE when v = 0, but not KE.
Answer:
Answer is 14.65
Refer below.
Explanation:
Refer to the picture for brief explanation.
Answer:
A
Explanation:
because the water carRies but also compacts
We know the equation
weight = mass × gravity
To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.
So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)
Therefore,
100N = mass × 10
mass= 100N/10
mass= 10 kg
Now, all we need are the moon's gravitational field strength and to apply this to the equation
weight = 10kg × (gravity on moon)