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nadezda [96]
3 years ago
12

You are pulling a 80 kg box with a rope. The force you exert on the box is 90 N at 30 degrees from the ground. What would be the

coefficient of kinetic friction if the box is moving at constant velocity?
Physics
1 answer:
Feliz [49]3 years ago
7 0

Answer:

Coefficient of dynamic friction= md= 0.09931

Explanation:

To determine the coefficient of dynamic friction we must first match the friction force that is permendicular to the normal force of the block and opposite to the drag force, to the component of the drag force in this same direction. This component on the X axis of the drag force will be:

F= 90N × cos(30°) = 77.9423N

This component on the X axis of the drag force must be equal to the dynamic friction force that is equal to the coefficient of dynamic friction by the normal force of the block weight:

F= md × m × g= 77.9423N

m= mass of the block

md= coefficient of dynamic friction

g= gravity acceleration

F= md × 80kg× 9.81 (m/s²)= 77.9423(kg×m/s²)

md= (77.9423(kg×m/s²) / 784.8 (kg×m/s²)) = 0.09931

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The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

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