Question

1. a book weighs 4 n. when held at rest in your hands, the net force on the book is

a. 0 n
b. 0.4 n
c. 4 n
d. 40 n

1. a tennis ball and a solid steel ball with the same diameter are dropped at the same time. which ball has the greater force acting on it?

a. the steel ball
b. the tennis ball
c. they both have the same force acting on them.

1. the reason a tennis ball and a solid steel ball will accelerate at the same rate, in the absence of air resistance, is that...

a. they have the same mass.
b. the ball with the larger force has the smaller mass.
c. the ball with the larger force also has the larger mass.
d. the force acting on them is the same.

1. suppose a cart is being moved by a force. if suddenly a load is dumped into the cart so that the carts mass doubles, what happens to the carts acceleration?

a. it quarters
b. it halves
c. it stays the same
d. it doubles

1. you pull horizontally on a 50 kg crate with a force of 450 n, and the friction on the crate is 250 n. the acceleration of the crate is...

a. ( 2 \frac { m } { s ^ { 2 } } )
b. ( 4 \frac { m } { s ^ { 2 } } )
c. ( 9 \frac { m } { s ^ { 2 } } )
d. ( 14 \frac { m } { s ^ { 2 } } )

1. the diagram below shows a 4 kg object accelerating at ( 10 \frac { m } { s ^ { 2 } } ) on a rough horizontal surface.

acceleration ( = 10 mathrm { m } / mathrm { s } ^ { 2 } )
what is the magnitude of the frictional force, ( f _ { f } ), acting on the object?
a. 5 n
b. 20 n
c. 10 n
d. 40 n

Explanation:

Step1: Apply Newton's second law

Newton's second law is \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. The net force \(F_{net}\) on the object is given by \(F_{net}=F_{applied}-F_{f}\) (where \(F_{applied}\) is the applied force and \(F_{f}\) is the frictional force).

Step2: Substitute into the formula

We know \(m = 4.0\space kg\), \(a=10\space m/s^{2}\), and \(F_{applied}=60\space N\). From \(F_{net}=ma\), we have \(F_{applied}-F_{f}=ma\). Rearranging for \(F_{f}\), we get \(F_{f}=F_{applied}-ma\).

Step3: Calculate the frictional force

Substitute the values: \(F_{f}=60 - 4\times10\). First, calculate \(4\times10 = 40\). Then \(F_{f}=60 - 40=20\space N\).

Answer:

B. \(20\space N\)