Question

1. The force    = 9 i + 12 j   + 5 k   acts on a small object. If the displacement of the object is    = 16 i + 10 j . Find the work done by the force using the scalar product .

2. A 25 kg chandelier is hanging from a cable in an elevator (have you ever seen a chandelier in an elevator?). What is the tension in the cable if the elevator that moves downward with an acceleration of 3.2 m/sec2?

3. A car travels around a 75 m radius flat curve at a constant speed. If the coefficient of friction which allows the car to travel around the curve is 0.8, what is the maximum speed that the car can go around the curve without skidding? (The mass of the car is m.)

4. A block of mass m lies on a plane as shown below. The coefficient of friction between the block and the plane is _uk = 0.2. Determine the acceleration of the block as it slides down the incline. 30 degrees is the angle.

5. Two crates, of mass 100 kg and 40 kg, are in contact with each other on a horizontal surface as shown below. The surface has u_k = 0.4. A 1250 N force is exerted on the 100 kg crate. Calculate the acceleration and the contact force between the two crates.

6. Calculate the speed of a satellite placed two earth radii above the earth’s surface.

(m_Earth = 5.98 x 10²⁴ kg, r_earth = 6.38 x 10⁶ m)

7. A 28 kg block is connected to an empty 2 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.45 and the coefficient of kinetic friction between the table and the block is 0.32. Sand is gradually added to the bucket until the system just begins to move. Calculate the mass of sand added to the bucket.

Answer 1

I. Z - q2 + 122+ 5 2 = 162+ lo Work done w = Fod = 9i+12h+SÊ. (169+109) = 9x16+ 2x10 +5x0 -1264 2. T-ig --ma - T=ng-ma ing =9579.8-25*3.2 = 25*(9.8-3.2 = 25x 6.6 = 165N Hence tension [165 N 3. R = 75m, M = 0.8 Let maximum speed av mr² ung Rcontripetal force friction force = = Jugr = 10.8x9.8875 KE 24-25 mi's
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N 4. gy 67 ya 30° pr ing 300 x let acceleration is a - Efx mg sin 30 - f = ma- Ety N-ing as 30° = N = ng cos 30° Now f = MN = ung cas30° put in egna ung sin 30º- emg cos 30º ema a) a = g sin 30 - MG Cas30° =) a = 9.8 x-0.28%8* 3 13.2 m/s² S. 1250N tookg Gorg f of secu(100+4019 = 0.4 x 1409.8 - 548,8 1250-548.8 5 mpg2 100+40

Now contact force between the blocks 1350 yo mgE N fall MX1ooxg -0.4*100X9.8 = 392 - 1250-f-N = 100 a => 1250-392 -N = 100x5 1250-302-500 N -> N = 358N Hence acceleration 5 m/s2 contact force between the blocks 1358 N 6 GMM ma 7 72 V= /GM 7 here = R+ 2 R = 3R =) V = GM 16.678101245.988/029 BR 6.38x106 = v= 2500ml speed og sattelite = 2500 m/s

7. 28kg sand of mass 2 ем in kg bucket = 2kg Freebody of bucket & sand AT T= (m+2) g (m + 2) 7 free body diagram of block ta ->T N 289 N = 288 = f = MN = 0.45X282 T=f -) (m+2) = 0.45x289 0.45X28-2 = 10.6 kg 10.6kg m mass of sand - 10