A LEVEL PROBLEMS

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1) A 220-kg load is lifted 21.0-m vertically with an acceleration a = 0.150 g (where g = 9.80 m/s²) by a single cable. Determine (a) the tension in the cable, (b) the net work done on the load, (c) the work done by the cable on the load, (d) the work done by gravity on the load, (e) the final speed of the load assuming that it started from rest.

SOLUTION:

(a) From the force diagram we write SF_y = ma_y:

F_T – mg = ma;

F_T – (220 kg)(9.80 m/s²) = (220 kg)(0.150 m/s²),

which gives F_T =2.48 ´ 10³ N.

(b) The net work is done by the net force:

W_net = F_neth = (F_T – mg)h

= [2.48 ´10³ N – (220 kg)(9.80 m/s²)](21.0 m) = 6.79 ´10³ J.

(c) The work is done by the cable is

W_cable = F_Th

= (2.48 ´10³ N)(21.0 m) =5.21 ´10⁴ J.

(d) The work is done by gravity is

W_grav = – mgh

= – (220 kg)(9.80 m/s²)(21.0 m) =– 4.53 ´10⁴ J.

Note that W_net = W_cable + W_grav.

(e) The net work done on the load increases its kinetic energy, so use the work-kinetic energy theorem.

W_net = Dk = ½mv² – ½mv₀² ;

6.79 ´10³ J = ½ (220 kg)v² –0, which gives v =7.86 m/s.

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2) What should be the spring constant k of a spring designed to bring a 1200-kg car to rest from a speed of 100 km/h so that the occupants undergo a maximum acceleration of 5.0 g?

SOLUTION:

The maximum acceleration will occur at the maximum compression of the spring:

kx_max = Ma_max= M(5.0g), which gives x_max = 5.0Mg/k.

For the motion from reaching the spring to the maximum compression of the spring, 

we use energy conservation: 

k_i + U_springi = k_f + U_springf;

½Mv² + 0 = 0 + ½kx_max².

When we use the previous result, we get

½Mv² = ½k(5.0Mg/k)², which gives

k = 25Mg²/v² = 25(1200 kg)(9.80 m/s²)²/[(100 km/h)/(3600 s/h)]² = 3.7 ´10³ N/m.

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3) A 0.520-kg wooden block is firmly attached to a spring (k = 180 N/m). It is noted that the block-spring system, when compressed 5.0-cm and release, stretches out 2.3-cm beyond the equilibrium position before stopping and turning back. What is the coefficient of kinetic friction between the block and the table?

SOLUTION:

On the level the normal force is F_N= mg, so the friction force is F_fr = µ_kmg.

The block is at rest at the release point and where it momentarily stops before turning back.

For the work-energy principle, we have

W_NC = Dk + DU = (½mv_f² – ½mv_i²) + (½kx_f² – ½kx_i²);

–µ_kmg L = (0 – 0) + ½k(x_f² – x_i² );

– µ_k(0.520 kg)(9.80 m/s²)(0.050 m + 0.023 m) = ½ (180 N/m)[(0.023 m)² – (– 0.050 m)²],

which gives µ_k =0.48.

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4) A 1000-kg car has a maximum power output of 120 hp. How steep a hill can it climb at a constant speed of 70 km/h if the frictional forces add up to 600 N?

SOLUTION:

From the force diagram for the car, we have:

x-component: F – F_fr - mg sin q = 0

Because the power output is P = Fv, we have

(P/v) – F_fr= mg sin q.

The maximum power determines the maximum angle:

(P_max/v) – F_fr= mg sin q_max ;

(120 hp)(746 W/hp)/[(70 km/h)/(3/6 ks/h)] – 600 N = 

(1000 kg)(9.80 m/s²) sin q_max , 

which gives sin q_max = 0.409,orq_max = 24°.

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5) A small particle drops onto a spring as shown in the diagram. It compresses the spring a distance of 0.25 m before coming to a stop. Calculate the spring constant of the spring.

SOLUTION:

The energy of the mass is purely potential before it was dropped. 

E_i = mgy_i = (0.10kg)(9.80 m/s²)(1.0 m) = 0.98 J

The energy is purely potential once the spring is fully compressed, however it has both gravitational potential and elastic potential energy.

E_f = mgy_f + ½ ky²_f = (0.10 kg)(9.80 m/s²)(-0.25 m) + (0.5)k(-0.25 m)² = -0.245 J + .03125 k

Conservation of energy applies:

0.98 J = -0.245 J + .03125 k

k = 39.2 N/m

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