# 普通物理的問題! ((運動學

1.

A 40.0-kg child swings in a swing supported by two chains, each 3.00 m long. If the tension in each chain at the lowest point is 350 N, find (a) the child's speed at the lowest point and (b) the force exerted by the seat on the child at the lowest point. (Neglect the mass of the seat.)

Solution

(a) v =____m/s

(b) N =____N

2.

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is

391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person's mass, and (c) the acceleration of the elevator.

Solution

(a) W =____N

(b) m =____kg

(c) a =____m/s2

6.

A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.00 m/s. The coefficient of kinetic friction between sled and ice is 0.100. Use energy considerations to find the distance the sled moves before it stops.

Solution

△x =____m

10.

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.00 ´104 m/s. What will its speed be when it is very far from the Earth? Ignore friction and the rotation of the Earth.

Solution

vf =____ × 104 m/s

### 1 個解答

• 1 0 年前
最佳解答

1.

ΣFy = T + T - mg = ma = mv^2 / r

=> v^2 = r (2T – mg) / m = (3) [2(350) – (40)(9.8)] / 40 = 23.1

=> v = 4.81 m/s

F = N - mg = m v²^2/ r

=> N = m v² / r + mg = 2 T = 700 N

2.

令電梯加速度為a.

591=m(9.8+a) --(1)

391=m(9.8-a) --(2)

m=50 kg

a=2 m/s2

W=50*9.8=490 N

6.

(1/2)mv^2=µmgh

h=(1/2)(4)/(0.1*9.8)=2.04 m.

10.

-mMG/R+(1/2)mv^2=0+(1/2)mV^2

=>-MG/R+(1/2)v^2=(1/2)V^2

=>-(6.67*10^-11)(5.98*10^24)/(6.37*10^6)+(1/2)(2*10^4)^2=(1/2)V^2

=>-6.26*10^7+2*10^8=0.5V^2

=>V=1.66*10^4 m/s