6) a 1500kg car is travelling along a straight road at 20m/s. two sec.later its

Question

6) a 1500kg car is travelling along a straight road at 20m/s. two sec.later its speed is 21m/s.what is the magnitude of the net force acting on the car during this time.

7) a 2.0kg book.is lying on a 0.75 m high table.you pick it up and place it on a bookshelf 2.3 m above the floor.
a) how much work does gravity fo on the book
b) how much work does your hand do on the book?

8) you are sitting in your physic vlass next to another student.60 m away.estimate the magnitude of the gravitational force between you.assume you each have a mass of 80 kg.
(G= 6.67x10^_11N_m^2/kg^2)

9) a car is travelling at 10m/s
a) how fast would the car need to go to double its kinectic energy
b)by what factor does the car s kinectic energy increase it its the speed is double to 20m/s?

10) how much energy can be stored in a spring with a spring constant of 500M/m if its maximum possible stretch is 20cm?

10) bob can throw a 500g rock with a speed of 30m/s.he moves his hand forward 1.0m while doing so.
a) how much gorce, assume to be constant, does bob apply to the rock?

b) how much work does bob do on the rock?

Explanation / Answer

6) his involves two parts, first it's a constant accleration problem:
u = 20 m/s
v = 21 m/s
s = ?
a = ?
t = 2 s

v = u + at
21 = 20 + 2a
1 = 2a
a = 1/2
a = 0.5 m/s^2

Second, you have to know how to use Newton's second law:
F = ma = 1500 * 0.5 = 750 N

7) Part A:
Work = Force * distance
In this problem, the force is the weight of the book.
Weight = 2 * 9.8 = 19.6 N
To determine the distance subtract the two heights.
d = 2.3 – 0.75 = 1.55 m
Work = 19.6 * 1.55 = 30.38 N * m
Since the direction of the book’s weight is opposite the direction the book is moving, the work is negative.

Part B:
To lift the book, your hand must exert a force that is equal to the book’s weight. Since you lift the book 1.55 meters, the work that you hand does on book is 30.38 N * m. Since the direction that your hand is moving is the same as the direction the book is moving, the work is positive.

8) F = Gm1m2/r^2 = (6.67*10^-11*80*80)/0.60^2 = 1.186*10^-6 N

9) a) KE = 1/2 m V^2
Since the first part (1/2 m) doesn't change, you need a new value for V such that

V(new)^2 = 2 * V(old)^2
V(new) = (2 * 10^2) ^ 1/2
V(new) = 200^1/2
which is approx. = 14.14 meters per second.

b)

Old KE = 0.5*m*100 = 50m J

New KE = 0.5*m*400 = 200m J

Ratio = 4 times more

10) K.E. = 0.5kx^2.........k is spring constant and x is the stretch.

K.E. = 0.5*500*0.20^2 = 10.0 J

11) b) Work = E = ½mv² = ½ * 0.5kg * (30m/s)² = 225 J

a) F = W / d = 225J / 1m = 225 N

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.