What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?
1 answer:
Answer:
Fr = 48 [N] forward.
Explanation:
Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.
∑F = Fr
The movement remains forward, since the force produced by the movement is greater than the braking force.
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(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:
where
m = 62.0 kg is the mass of the passenger
is the change in velocity of the car (and the passenger), which is
So, the linear impulse experienced by the passenger is
(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:
where in this case
is the linear impulse
is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:
The angular speed of the device is 1.03 rad/s.
<h3>What is the conservation of angular momentum?</h3>A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.
Using the conservation of angular momentum
Here, = the system's angular momentum before the collision
= 0 + mv
= (0.005)(450)(0.752)
= 1.692 kgm²/s
The moment of inertia of the system is given by
I = 2(M₁R₁² + M₂R₂²)+ mR₁²
= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²
= 1.6292 kgm²
Here, = Iω
So,
1.692 = 1.6292(ω)
ω = 1.03 rad/s
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