Question

Q1)Two forces, F⃗ 1 and F⃗ 2, act at a point. The magnitude of F⃗ 1 is 9.80 N , and its direction is an angle 58.0 ∘ above the x-axis in the second quadrant. The magnitude of F⃗ 2 is 6.40 N , and its direction is an angle 52.7 ∘ below the x-axis in the third quadrant.

a-What is the x-component of the resultant force?

b-What is the y-component of the resultant force?

c-What is the magnitude of the resultant force?

----------------------------------------------------------------------

Q2)A crate with mass 32.0 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 120 N .

a-What acceleration is produced?

b-How far does the crate travel in 10.0 s ?

c-What is its speed at the end of 10.0 s ?

----------------------------------------------------------------------

Q2)An electron of mass 9.11×10⁻³¹ kg leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is a distance 2.65 cm away. It reaches the grid with a speed of 4.10×10⁶ m/s . The accelerating force is constant.

a-Find the acceleration.

b-Find the time to reach the grid.

c-Find the net force. (You can ignore the gravitational force on the electron).

----------------------------------------------------------------------

Q4)At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 47.0 N at the surface of the earth. In this problem, use 9.81 m/s2 for the acceleration due to gravity on earth.

a-What is its mass on the earth's surface?

b-What is its mass on the surface of Io?

c-What is its weight on the surface of Io?

----------------------------------------------------------------------

Q5)The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N.

a-What is the magnitude of the acceleration?

----------------------------------------------------------------------

Q6)A chair of mass 16.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F = 38.0 N that is directed at an angle of 43.0 ∘ below the horizontal and the chair slides along the floor.

a-Use Newton's laws to calculate the normal force that the floor exerts on the chair.

----------------------------------------------------------------------

Q7)An athlete whose mass is 94.5 kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 540 N . He lifts the barbell a distance of 0.60 m in a time of 1.6 s .

a-Use Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell.

Express your answer using two significant figures.

----------------------------------------------------------------------

Answer 1

Q1 .

a) x-component of resultant force = F1 Cos(58) + F2 Cos(52.7)

= 9.8 * 0.53 + 6.4 * 0.605

= 9.06 N

b) y-component of resultant force = F2 Sin(52.7) - F1Sin(58)

= 6.4 * 0.795 - 9.8* 0.848

= - 3.222 N

c) Resultant force = = 9.615 N

Q2.

mass = m = 32kg

Force = F = 120N

a) acceleration = F/m = 120/32 = 3.75 m/s^2

b)

initial speed = u = 0

time = t = 10s

acceleration = a = 3.75

distance = ut + 1/2 a t^2

= 0 + 1/2 * 3.75 * 100 = 187.5

Distance travelled in 10 seconds = 187.5 m

c) Final speed = v , time = t = 10s

v = u +at

=0 + 3.75 * 10 = 37.5 m/s^2

Q3

a) v^2 = u^2 + 2 * a * s

(4.1*10^6)^2 = 0 + 2 * a * (0.0265)

a = (4.1 * 10^6)^2 / 2*0.0265  

= 3.1 * 10^14 m/s^2

b) v= u+at

4.1* 10^6 = 0 + 3.1 * 10^14* * t

t = 4.1 * 10^6 / 3.1* 10^14 = 1.3 * 10^-8

Time taken = 1.3 * 10^-8 seconds

c) Net force = mass * acceleration

= 9.11 * 10^-31 * 3.1 * 10^14 = 2.8 * 10^-16 N

Net force = 2.8 * 10^-16 N