Question

1. a large banana crate, with a mass of 75.0 kg, is pushed across the warehouse floor by a thieving chimpanzee at a constant velocity. what force must be exerted by the chimp in order to keep the crate moving at the same velocity? you are given that μₖ = 0.25 between the crate and the floor.

a) 19 n b) 1.8×10² n c) 3.0×10² n d) 7.4×10² n

1. an airplane moves in straight level flight at constant velocity. what is the net force acting on it if the mass of the plane is 1000 kg and the frictional drag of the air is 1800 n?

a) 0 n b) 1800 n c) 10 000 n d) 11 800 n

1. when the space shuttle lifts off the pad at cape kennedy space centre, its engines provide a total thrust of 3.4 x 10⁷ n. if the shuttle has a mass of 2.2 x 10⁶ kg, what is its initial acceleration?

a) 1.5 m/s² b) 5.6 m/s² c) 9.8 m/s² d) 7.4×10²
(image of space shuttle)

1. half way through, time for a brain break. now, a good estimate of the number of students taking this final exam are:-

i’m not supposed to look around just incase i’m accused of cheating! (assume all students were present).
a) 10! b) 20! c) 30! d) 40!

1. the frictionless system shown below accelerates at 5.70 m/s² when released.

(image of a frictionless table with a block m and a 2.00 kg mass hanging)
find the tension in the string while the system is accelerating.

Explanation:

Question 27

Step1: Identify forces (constant velocity)

Since velocity is constant, net force \( F_{net} = 0 \). So applied force \( F_{app} \) equals kinetic friction \( F_{k} \).

Step2: Calculate normal force \( F_N \)

For horizontal surface, \( F_N = mg \), where \( m = 75.0\,\text{kg} \), \( g = 9.8\,\text{m/s}^2 \).
\( F_N = 75.0 \times 9.8 = 735\,\text{N} \).

Step3: Calculate kinetic friction \( F_k \)

\( F_k = \mu_k F_N \), \( \mu_k = 0.25 \).
\( F_k = 0.25 \times 735 = 183.75\,\text{N} \approx 1.8 \times 10^2\,\text{N} \).

Step1: Recall Newton’s 1st law

Constant velocity means acceleration \( a = 0 \). By \( F_{net} = ma \), \( F_{net} = 1000 \times 0 = 0\,\text{N} \).

Step1: Identify forces (thrust, weight)

Thrust \( F_T = 3.4 \times 10^7\,\text{N} \), weight \( F_g = mg \), \( m = 2.2 \times 10^6\,\text{kg} \), \( g = 9.8\,\text{m/s}^2 \).
\( F_g = 2.2 \times 10^6 \times 9.8 = 2.156 \times 10^7\,\text{N} \).

Step2: Net force \( F_{net} = F_T - F_g \)

\( F_{net} = 3.4 \times 10^7 - 2.156 \times 10^7 = 1.244 \times 10^7\,\text{N} \).

Step3: Acceleration \( a = \frac{F_{net}}{m} \)

\( a = \frac{1.244 \times 10^7}{2.2 \times 10^6} \approx 5.65\,\text{m/s}^2 \approx 5.6\,\text{m/s}^2 \).

Answer:

B) \( 1.8 \times 10^2\,\text{N} \)

Question 28