A person in the back of a pick-up moving at 20 m/s relative to the Earth, throws a baseball with a speed of 15 m/s in a directio
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The speed of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.
We are given that-
the mass of the elevator (m) = 1000 kg ;
the distance the elevator decelerated to be y = 8m ;
the tension is T = 11000 N;
let us determine the acceleration 'a' by using Newton's second law of motion.
∑Fy = ma
W - T = ma
(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a
9800 - 11000 = 1000
a = - 1.2 m/s²
Using the equation of kinematics to determine the initial velocity.
² =
² + 2ay
= √ ( 2 x 1.2m/s² x 8 m )
= √19.2 m²/s²
= 4.38 m/s ≈ 4 m/s
Hence, the initial velocity of the elevator is 4m/s.
Read more about the Equation of kinematics:
#SPJ4
Explanation:
Given that,
The slope of the ramp,
Mass of the box, m = 60 kg
(a) Distance covered by the truck up the slope, d = 300 m
Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :
u and v are the initial and the final velocity of the truck
(b) The work done on the box by the force of gravity is given by :
Here,
W = -24550.13 J
(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.
(d) The work done by friction is given by :
Hence, this is the required solution.
KE =
Convert g in kg
Find v:
2KE = mv^2
v=√2KE/m
v=√2*31.5/0.00245
v=160 m/s
Convert g in kg
Find v:
2KE = mv^2
v=√2KE/m
v=√2*31.5/0.00245
v=160 m/s