Net force?

Question

Always T or can be F.

A) During the collision of a car with a large truck, the car exerts a lesser force on the truck than the truck exerts on the car.

False, equal and opposite forces.

B) The net force which acts on an object which maintains a constant velocity is zero.

True, if F=ma and (a) due to constant velocity = 0, then F=0.

C) If a net force acts on an object, the object's speed will change.

False, what if the F is downward on a still box? The speed is the same but the F is different.

D) If two objects have the same acceleration, they are under the influence of equal forces.

False, one object could have a greater mass, therefore a larger force would have to be applied to get the same accel.

E) A bicycle initially moving at a constant velocity will slow down unless a small net force is applied.

I don't know what this question is saying


F) An object's velocity will change if a net force acts on the object

False? Same reason as a still box?

Answer 1

E) True, if it is assumed that frictional forces and aerodynamic drag are not zero. If there are no forces that retards motion, then the bycycle will not slow down.

F) True. If a nonzero net force acts on a body, the velocity of the object will change in accordance with F=ma.

You answer for C) is indeed correct, but your explanation isn't very good so I don't quite know if you caught the subtlety of the question. The difference between C) and F) is that C) says the object's SPEED will change while F) says the object's VELOCITY will change while force acts on the object. Speed is not a vector. Velocity is. The law of motion only says the velocity will change under a force. It says nothing about speed. It is entirely possible for an object to be under a force, be accelerated, change velocity, and yet keep the speed constant. This is exactly what happens in uniform circular motion.

Answer 2

First of all I think you do not understand what NET FORCE is. The NET FORCE is the VECTORIAL SUM of ALL FORCES acting on the object. So if the sum of the forces acting on the object is ZERO, that object will be in constant movement.
Think about it: if a body moves at a constant speed on a horizontal plane, it will be subjected to:
GRAVITY PULL: G = m * g (gravitational acceleration 9.8 m/s2)
FRICTION: Ff = G * fk (friction coefficient)
REACTION FORCE OF THE PLANE OF MOVEMENT:
R = G
MOVEMENT FORCE: Ft - traction force exerted upon the body.
If the NET FORCE (that is the sum of all these) is zero, then the body will maintain its preexistent velocity. Be it zero or more than zero - movement or immobility.
One should always think about THE SUM of forces as the NET FORCE. So if the NET FORCE is zero, the object does not modify its state of movement. If it is different from zero, it does. If the NET FORCE is directed vertically up it will diminish its weight and thus diminish the friction and cause it to accelerate. If the NET FORCE exceeds gravity it will cause the object to also change its trajectory. If it is directed DOWNWARDS, it will increase the friction force and cause the object to slow down.
Same with various degrees of BACK or FORWARD NET FORCES. Once again, we are speaking about the RESULTANT OF ALL FORCES ACTING ON THE OBJECT! Therefore, B is true, C is also TRUE, D is obviously FALSE (TRUE only if objects have same masses and friction coefficients).E is also FALSE. F is TRUE. Of course, we have chosen to ignore the aerodynamic resistance, but since it can be equivalent (to a certain extent) to a friction force I think it is reasonable enough.

Answer 3

Great job asking this one. The bike will slow down because the arrow of gravity must be included. (I should say vector, but I say arrow) and the arrow of friction should be included.
The objects velocity will change can be still box ifvelocity is zero. True if velocity equals zero. If v does not equal zero, remember to include negative acceleration of friction that must be overcome.

Answer 4

(c)Your ans is wrong coz if a net force acts on a body,it is
accelerated and its speed changes.
In the case about which you are saying ,net force on the body is still zero as normal rxn. will balance the force of gravity.

(E)Bicycle will not slow down unless a force is applied.

(F)True(clarified above)


Rest of the answers are absolutely correct

Answer 5

Friction is definitely the biggest. on your first equation you will desire to define the path for the formulation to be valid. so which you would be able to desire to state. "interior the path of holiday F(internet) = F(engine) - F (friction) in any respect situations while the engine is making use of any forward rigidity. If the motor vehicle is slowing down then F (friction ) is larger than F (engine ) and F(internet) is unfavorable" that's a finished assertion no longer open to misinterpretation. throughout deceleration that's particularly achieveable that the formulation would desire to be F(internet) = F(brakes) + F (engine) - F (friction) the place F (brakes) and F (engine) are the two unfavorable in the present day. Your assumption is that the engine is performing interior the different path to the friction. i.e utilising forward. that's often no longer valid at parkway speeds while the foot is removed from the accelerator. there are a number of varieties of friction. What your instructor is calling for would selection. you would be able to desire to specify a number of them in case you needed to. surprisingly for section 2 i.e air resistance, rolling resistance of the tyres and the drag of the engine. (confident there is drag at those speeds. The engine applies a utilising rigidity no count if that's slowed to below its idle speed, besides the undeniable fact that it applies resistance no count if that's speeded up, via some exterior rigidity, to greater advantageous than its idle speed that explains your confusion over the automated automobile on the site visitors lights fixtures the engine is slowed via the torque convertor so that's attempting to hurry up back to its idle speed. Even at larger speeds this situation is genuine until the value of the motor vehicle exceeds the optimum speed ratio of your particular torque convertor. This varies from make and kind of motor vehicle.)