Solved: Question 21 A Constant Net Force On A Rail-Road Car Produces Constant

Force may be defined as the cause of motion and deformation.

Đang xem: A constant net force on a rail-road car produces constant

When a force is applied to an object, the object either moves or changes shape or both. In most cases, it is not possible to detect the deformation by naked eyes at the molecular or atomic level. Deformation occurs no matter how small.

In Chapter 1, force was defined as the product of mass and acceleration. Simply F = Ma. A more useful form of this formula is ΣF = Ma. ΣF means the sum of forces acting on mass M. Since forces acting on an object may act in opposite directions, ΣF is also called the net force.

The formula ΣF = Ma is called the “Newton”s 2nd Law of Motion.”

For example, a car moving along a straight and horizontal highway, experiences an engine force Fe, while being opposed by an overall frictional force, Ff ( road friction as well as air resistance). If the car is moving to the right and to the right is taken to be the positive direction, thenFe acts to the right and Ff acts to the left. The net force isΣF =Fe – Ff.

Example 1: An 850-kg car is accelerating at a rate of 2.4m/s2 to the right along a straight and horizontal road where it experiences an overall frictional force of 1500N. Determine the force that its engine exerts.

Solution: If “to the right” is taken to be positive as usual, Fe is positive, and Ff negative. Note that friction always opposes the direction of pending motion. Applying Newton”s 2nd law:

ΣF = Ma; Fe – Ff = Ma ; Fe – 1500N = (850kg)(2.4m/s2) ;Fe = 1500N + 2040N =3500N

Note that the 2040N must be rounded to 2 significant figures and then added to the 1500N.

Example 2: A 2400-kg truck is moving at a constant speed of 15m/s on a horizontal and straight road that offers an overall frictional force of 1800N. Calculate (a) its acceleration, (b) the engine force, (c) the distance it travels in 35s, (d) its acceleration if it changes its speed to 25m/s in 8.0 seconds, and (e) the engine force in this case.

Solution: (a) Since the truck”s speed is initially constant; therefore, a1 = 0.

(b) ΣF = Ma; Fe – Ff = Ma1 ; Fe – 1800N = (2400kg)(0) ;Fe = 1800N + 0 =1800N

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(c) x = v t (Equation of motion forconstant velocity) ; x = (15m/s)(35s) = 530m (rounded to 2 sig. fig.)

(d) a2 = (Vf – Vi) / t ; a2 = (25m/s – 15m/s) / 8.0s = 1.25 m/s2

(e)ΣF = Ma; Fe – Ff = Ma2 ; Fe – 1800N = (2400kg)(1.25m/s2) ;Fe = 1800N + 3000N =4800N

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Example 3: A car that weighs 14700N is traveling along a straight road at a speed of 108 km/h. The driver sees a deer on the road and has to bring the car to stop in a distance of 90.m. Determine (a) the necessary deceleration, (b) the stopping force, (c) the brakes force, if the road friction is 2100N, and (d) the stopping time. Try to solve the problem yourself before looking at the solution.

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Solution: The mass of the car and its velocity in (m/s) must be determined first. Recall that w = Mg. The mass of the car is therefore, M = w/g, or M = (14700N) / (9.8 m/s2) , orM = 1500kg.

Vi = (108 km/h ) (1000m / 1km) ( 1h / 3600s) = 30.m/s(Use horizontal fraction bars only)

(a) Vf2 – Vi2 = 2 a x ; (0) 2 – (30.)2 = 2 (a) (90.0m) ;-900 = 180a ;a = – 5.0 m/s2 (deceleration).

(b)ΣF = Ma ;ΣF = (1500kg)(- 5.0 m/s2 ) = -7500N

(c)ΣF = Fbrakes + Ffriction ; -7500N = Fbrakes- 2100N ; Fbrakes = – 5400N.

(d) a = (Vf – Vi) / t ; t = (Vf – Vi) / a ; t = (0 – 30.m/s) / (-5.0m/s2) ;t = 6.0 seconds

Newton”s Laws:

1) If an object is under a zero net force, it is either stationary or if moving, it moves at constant velocity. Note that constant velocity means constant speed plus constant direction that means along a straight line.

2) A nonzero net forceΣF acting on mass M causes an acceleration a in it such thatΣF= Ma. The acceleration has the same direction as the applied net force.

3) There is a reaction for every action, equal in magnitude, but opposite in direction.

An example covering the 1st and 2nd laws has already been made (Example 2). For the 3rd law, look at the following example:

Example 4: A 20.0-kg crate is on a horizontal and frictionless surface as shown. (a) Calculate and show the vertical forces acting on this crate. (b) Knowing that the crate is being pushed to the left by a 53-N force, what magnitude force (F) to the right must be applied onto the block to give it an acceleration of 2.5m/s2to the right?

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Note that w = 196 N must be rounded to 2 sig. fig. and written as w = 2.0x102 N. The same is true for F = 103N that must be rounded to 100 and written as F = 1.0x102 N.

Example 5: An 80.0-kg man is standing in an elevator. Determine the force of the elevator onto the person if the elevator is (a) accelerating upward at 2.5m/s2, (b) going upward at constant speed, (c) coming to stop going upward at a deceleration of 2.5m/s2, and (d) going downward at an acceleration of 2.

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5m/s2.

Solution: The force of the elevator onto the person is nothing but the normal reaction,N, of the floor onto his feet. For each case, a force diagram must be drawn. Let”s take the +y-axis to be upward. Make sure that you carefully draw each force diagram with minimal looking at the following force diagrams.

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