Answer:
1. The work done on the cube during the time the cube is in contact with the spring is 0.8023705 J
2. The speed of the cube at the instant just before the sliding cube leaves the ramp is approximately 31.5 cm/s
Explanation:
The given parameters of the Rube Goldberg machine are;
The distance from the free end of the spring to the top of the ramp, d = 17.0 cm = 0.17 m
The mass of the small cube to be launched, m = 119.0 g = 0.119 kg
The spring constant of the spring, k = 461.0 N/m
The angle of elevation of the ramp to the horizontal, θ = 50.0°
The coefficient of static friction of the wood, = 0.590
The coefficient of dynamic friction of the wood, = 0.470
The velocity of the cube at the top of the ramp, v = 45.0 cm/s = 0.45 m/s
The amount by which the cube is compressed, x = 5.90 cm = 0.059 m
The work done on the cube during the time the cube is in contact with the spring = The energy of the spring, E = (1/2)·k·x²
∴ E = (1/2) × 461.0 N/m × (0.059 m)² = 0.8023705 J
The work done on the cube during the time the cube is in contact with the spring= E = 0.8023705 J
2. The frictional force, =
·m·g·cos(θ)
∴ = 0.470 × 0.119 × 9.8 × cos(50) ≈ 0.35232 N
The work loss to friction, W = × d
∴ W = 0.35232 N × 0.17 m ≈ 0.05989 J
The work lost to friction, W ≈ 0.05989 J
The potential energy of the cube at the top of the ramp, P.E. = m·g·h
∴ P.E. = 0.119 kg × 9.8 m/s² × 0.17 m × sin(50°) ≈ 0.151871375 J
By conservation of energy principle, the Kinetic Energy of the cube at the top of the ramp, K.E. = E - W - P.E.
∴ K.E. = 0.8023705 J - 0.05989 J - 0.151871375 J ≈ 0.590609125 J
K.E. = (1/2)·m·v²
Where;
v = The speed of the cube at the instant just before the sliding cube leaves the ramp
∴ K.E. = (1/2) × 0.119 kg × v² ≈ 0.590609125 J
v² ≈ 0.590609125 J/((1/2) × 0.119 kg) ≈ 9.92620378 m²/s²
v = √(9.92620378 m²/s²) ≈ 3.15058785 m/s ≈ 31.5 cm/s
The speed of the cube at the instant just before the sliding cube leaves the ramp, v ≈ 31.5 cm/s.
Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature
We need to calculate the carrier lengthen
Using formula of length
Put the value into the formula
Hence, The carrier lengthen is 0.08436 m.
Answer:
the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Explanation:
The torque is given by :
where ;
m = 0.160 A.m²
B = 0.0800 T
θ = 35°
So the magnitude of the torque N = mBsinθ
N = (0.160)(0.0800)(sin 35°)
N = 0.007341
N = 7.34×10⁻³ Nm
Hence, the magnitude of the torque on the permanent magnet = 7.34×10⁻³ Nm
b) The potential energy
U = -mBcosθ
U = (- 0.160)(0.0800)(cos 45)
U = -0.010485
U = -1.0485 ×10⁻² J
Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils = -1.0485 ×10⁻² J
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
