Answer:
The maximum displacement ,A= 0.047 m
Explanation:
Given that
Mass, m = 45 g= 0.045 kg
spring constant ,K= 240 N/m
Initial velocity at equilibrium ,v= 3.5 m/s
Lets take the maximum displacement is A
As we know that when the mass reached at the extreme position then the velocity of the mass will become zero.
From energy conservation
Now by putting the values
A=0.047 m
The maximum displacement ,A= 0.047 m
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
Answer:
W = 34.64 ft-lbs
Explanation:
given,
Horizontal force = 4 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
W = F.d cos θ
W = 4 x 10 x cos 30°
W = 40 x 0.8660
W = 34.64 ft-lbs
Hence, work done on the box is equal to W = 34.64 ft-lbs
Answer:
h = 51020.40 meters
Explanation:
Speed of the rifle, v = 1000 m/s
Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :
h = 51020.40 meters
So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.
