Answer:
Explanation:
The energy stored in a spring
= 1/2 k x²
where k is spring constant and x is extension in the spring.
= .5 x 500 x .05²
= .625 J
Work done by friction = energy dissipated
= - μmg x d , μ is coefficient of friction , m is mass , d is displacement
= - .35 x 2.8 x 9.8 x .05
= - .48 J
energy of the mass when it reaches equilibrium position
= .625 - .48
= .145 J
If v be its velocity at that time
1/2 m v ² = .145
.5 x 2.8 x v² = .145
v² = .10357
v = .32 m /s
32 cm /s
Answer:
maximum speed of the block is 3.14 m/s
Explanation:
given data
mass = 1.70 kg
stretch in the spring = 0.2 m
time take by block to come to zero t = 0.2 s
solution
we know that Time period of oscillation (T) that is express as
T = 2t ......................1
put here value
T = 2 (0.2)
T = 0.4 s
so here time period is express as
T =
................2
here k is spring constant of the spring so put here value
0.4 =
here k will be
k = 419.02 N/m
so we use here conservation of energy that is
Maximum kinetic energy = Maximum spring potential energy ............3
(0.5) m v² = (0.5) k x²
here v is maximum speed block
so put here value and we get
(1.70) v² = (419.02) (0.2)²
v = 3.14 m/s
so maximum speed of the block is 3.14 m/s
The time it will take him to fall can be found from:-3m = -(g*t^2)/2
Find that time, it's the time the horse will travel horizontal while the cowboy is falling.So the horizontal distance away is 10 m/s * t
Answer:
option A
Explanation:
given,
height of the drop of stone = 9.44 m
speed of the stone = ?
As the stone is dropped the energy of the stone will be conserved.
using conservation of energy.
Potential energy = Kinetic energy

v = 13.60 m/s
Hence, the correct answer is option A
Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N