A 6 kg block is sliding down a horizontal frictionless surface with a constant speed of 5 m/s. It then slides down a frictionles
1 answer:
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Answer:
Explanation:
From the question we are told that:
Force P=88Ib
Mass of crate M_c=210Ib
Generally the equation for Frictional force F is mathematically given by
with
Therefore since Static Friction supersedes applied force body remains at rest.
Frictional force =88Ib (negative)
Answer:
Impulse, |J| = 0.6716 kg-m/s
Force, F = 63.35 N
Explanation:
It is given that,
Mass of the baseball, m = 0.146 kg
Initial speed of the ball, u = 15.3 m/s
Final speed of the ball, v = 10.7 m/s
To find,
(a) The magnitude of this impulse.
(b) The magnitude of the average force of the glass on the ball.
Solution,
(a) Impulse of an object is equal to the change in its momentum. It is given by :
J = -0.6716 kg-m/s
or
|J| = 0.6716 kg-m/s
(b) Another definition of impulse is given by the product of force and time of contact.
t = 0.0106 s
F = 63.35 N
Hence, this is the required solution.
Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:
Since it is negative, the direction is clockwise.