Tutorial on Ch 3-5 Solution PDF

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Ch 3-5 Solutions...

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Tutorial Problems From Previous Tests and Exams 1 Tutorial Ch 3 - 5 (1) A test rocket is launched by accelerating it along a 200.0-m incline at 1.25 m/s2 starting from rest at point A (in the following figure.) The incline rises at 35.0 above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the maximum height above the ground that the rocket reaches, and (b) the greatest horizontal range of the rocket beyond point A. Solution: Once the rocket leaves the incline it moves in projectile motion. The acceleration along the incline determines the initial velocity and initial position for the projectile motion. For motion along the incline let x be directed up the incline.

vx2  v0 x2  2a x (x  x0 )

gives

v x  2(125 m/s2 )(200 m)  22 36 m/s.

When the projectile motion begins the rocket has v0  22 36 m/s at 350 above the horizontal and is at a vertical height of (2000 m)sin350  1147 m. For the projectile motion let x be horizontal to the right and let y be upward. Let y  0 at the ground. Then y0  1147 m, v0x  v0 cos35 0

2 m/s,

3 m/s, a x  0, a y   9 80 m/s 2. Let x  0 at point A, so

v0 y  v0 sin35 0

x0  (2000 m)cos35 0 163 8 m. 2

2

Solve: (a) At the maximum height we have v y  0. v y  v0 y  2a y ( y  y0 ) gives y  y0 

v y2  v0 y2 2a y



0  (12 83 m/s)2 2(9 80 m/s2 )

 8 40 m and

y  1147 m  840 m  123 m.

The maximum height above ground is 123 m. (b) The time in the air can be calculated from the vertical component of the projectile motion: y  y0  1147 m, 2

v0 y  1283 m/s, ay  980 m/s . y  y0  v0 yt  21 a yt 2 gives (4 90 m/s 2) t2 (12 83 m/s) t 1147 m. The quadratic formula gives

t





1 2 1283  (1283)  4(490)(114 7) s. The positive root is t  632 s. Then x  x 0  9 80

v0 x t  21 ax t 2  (1832 m/s)(632 s)  1158 m and

x  1638 m  1158 m  280 m. The horizontal range of the rocket is 280 m.

2

(2) A block (mass M = 0.25 kg) is at rest on a rough inclined plane (kinetic coefficient of friction µ k = 0.35 and angle θ = 30o) and is connected to an object with mass m = 0.40 kg which hangs freely, as shown. The rope may be considered massless; and the pulley may be considered massless and frictionless. Find the speed v of mass m when it has fallen distance y = 1.0 m.

M

m y

 Solution:

mg  T  ma T  k N  Mg sin  Ma N  Mg cos mg  k Mg cos   Mg sin   Ma  ma

N

v 2  v02  2ay

v

T a

fk=µ kN

g( m  M sin   k M cos a  ( M  m)

W  Fx  m.a .x 

T

m 2 2 (v  v 0 ) 2

2(9.8)(1.0)  0.40  (0.25)sin 30  0.35(0.25) cos30   2.45 m / s 0.25  0.40 

Mgsinθ Mgcosθ

mg

3

(3) A 5.0-kg block is placed on top of a 10.0 kg block as shown in Figure. A horizontal force of 45.0 N is applied to the 10.0 kg block, and the 5.0 kg block is tied to the wall. The coefficient of kinetic friction between all surfaces is 0.20 (a) Draw a free body diagram for each block. (b) Determine the magnitude of the acceleration of the 10.0-kg block.

5.0 kg 45.0 N 10.0 kg

Solution: U  upper block & L  lower block & LU  L on U T  Free-body diagram for each block  (fLU , fUL) and (NLU , NUL) appear in both diagrams as action-reaction pairs 5.0 kg block: Fy = 0 NLU = WU = mU g = (5.0)(9.8) = 49.0 N Fx = 0 T = fLU =  k mU g = (0.2)(5.0)(9.8) = 9.8 N 10.0 kg block: Fx = ma F  fUL  fTL = mLa (1) FUL Fy = 0 NTL  NUL  mLg = 0 (2) FTL = k NTL = k (NUL + mL g) = (0.2)(49.0 + 98.0) = 29.4 N FTL From Eq.(1), we get: Net force on L = 45  9.8  29.4 = (10.0)a Therefore, a = 5.8 / 10 = 0.58 m/s2 (b)

NLU FLU

5.0 kg

WU NTL NUL 10.0 kg

WL

F

4

(4) A penguin slides at a constant velocity of 1.73 m/s down an icy incline. The incline slopes above the horizontal at an angle of 10.0°. At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. How much time is required for the penguin to slide to a halt (stop) after entering the horizontal patch of ice?

θ =10.0o

Solution: The penguin comes to a halt on the horizontal surface because the kinetic frictional force opposes the motion and causes it to slow down. The time required for the penguin to slide to a halt (v = 0 m/s) after entering the v  v0 v0 horizontal patch of ice is t   ax ax We must, therefore, determine the acceleration of the penguin as it slides along the horizontal patch (see the following drawing).

5 For the penguin sliding on the horizontal patch of ice, we find from free-body diagram B and Newton's second law in the x direction (motion to the right is taken as positive) that

 Fx  – f k2  ma x

or

ax 

– f k2 m



– k FN2 m

In the y direction in free-body diagram B, we have  Fy  FN2 – mg  0 , or FN2  mg . Therefore, the acceleration of the penguin is

– k mg

 – k g (1) m Equation (1) indicates that, in order to find the acceleration ax, we must find the coefficient of kinetic friction. We are told in the problem statement that the coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. Therefore, we can use the motion of the penguin on the incline to determine the coefficient of friction and use it in Equation (1). For the penguin sliding down the incline, we find from free-body diagram A (see the previous drawing) and Newton's second law (taking the direction of motion as positive) that fk1  mg sin (2) or  Fx  mg sin  – fk1  max  0 Here, we have used the fact that the penguin slides down the incline with a constant velocity, so that it has zero acceleration. We know that fk1   k FN1 . Applying Newton's second law in the direction perpendicular to the incline, we have ax 

 Fy  FN1 – mg cos   0

or

FN1  mg cos 

Therefore, fk1  k mg cos , so that according to Equation (2), we find

fk1  k mg cos   mg sin  Solving for the coefficient of kinetic friction, we have

k 

sin   tan  cos

Finally, the time required for the penguin to slide to a halt after entering the horizontal patch of ice is

t t

v 0 ax



v0 –v 0  – k g g tan 

– v0 v0 v0 1.73 m/s     1.00 s ax – k g g tan  (9.80 m/s2 ) tan10...