physics既dynamics...20分!4題

[匿名]

2009-04-16 3:21 pm

1) It takes a force of 109N to list a stone straight up this force gives the stone an accleration of 12.0m/s2. Find the mass of the stone.

2)If a horizontial force of 30N is required to slide a 12.0kg wooden crate across the floor a constant velocity. What is the coifficient of sliding fricition between crate and floor?

3)A 70kg astronant is standing on a scale in a spaceship while the ship is moving in a straight line unil a constant velocity of 100m/s near a large planet, the scale reads 300N. The ship ther accelerates away from the planet at 7m/s2. What does the scale read now.

4)A 2500kg car is traveling at a constant speed of 14.0m/s along a icy, but straight and level road. The car seeing an approching traffic light turn red, starts on the breaks wheel locked and tires skidding the car slides to a halt in a distance of 25.0m. What is the cofficient of sliding friction between the tires and icy roadbed?

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回答 (1)

Prof. Physics

2009-04-16 10:35 pm

✔ 最佳答案

1. By Newton's 2nd law of motion,

F - mg = ma

102 - 10m = 12m

Mass, m = 4.64 kg

2. As the crate is moving with uniform velocity, the net force acting on it is zero.

So, friction = 30 N

Normal reaction, R = mg = 12 X 10 = 120 N

So, coefficient of sliding friction, u

= f / R

= 30 / 120

= 0.25

3. 300 = 70g

Gravitational field strength, g = 4.29 Nkg-1

By Newton's 2nd law of motion

F - mg = ma

F - 70(4.29) = 70(7)

Reading, F = 790 N

4. By equation of motion, v2 = u2 + 2as

0 = (14)2 + 2a(25)

Acceleration, a = -3.92 ms-2

By Newton's 2nd law of motion,

f = ma

f = 2500(3.92)

friction, f = 9800 N

Normal reaction, R = mg = 25 000 N

Coefficient of sliding friciton, u = f / R = 9800 / 25 000 = 0.392

參考: Physics king

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