Question

1. the free body diagram shows a block being pushed along a flat surface 98 n of normal force 10n of friction 200 n of pull 10 kg 98 n of gravitational force a) what direction and with how much net force is the block being pulled? 190n right b) what is the acceleration of the block? 190 = 10(a) a = 19 f = m(a) or a = f/m 19 define: momentum: 10) a 8 kg tire is rolled with a velocity of 25 m/s. what is the momentum of the tire? p = m(v) p = 8(25) the momentum of the tire was 200 kg(m)/s 11) a cyclist with a mass of 43 kg is riding a 7kg bicycle with a velocity of 5 m/s. what is the total momentum of the cyclist and bicycle? p = m(v) p = 43(5) + p = 43(7) the total momentum of the cyclist and bicycle was ____ kg(m)/s

Explanation:

Step1: Calculate net - force for part a

Net - force in the horizontal direction is the difference between the pulling force and the frictional force. The pulling force is $F_{pull}=200N$ and the frictional force is $F_{friction}=10N$. The normal force and gravitational force cancel each other out in the vertical direction. So, $F_{net}=F_{pull}-F_{friction}=200 - 10=190N$ to the right.

Step2: Calculate acceleration for part b

Use Newton's second law $F = ma$, where $F$ is the net - force, $m$ is the mass, and $a$ is the acceleration. Given $F = 190N$ and $m = 10kg$. Rearranging the formula for acceleration gives $a=\frac{F}{m}=\frac{190}{10}=19m/s^{2}$.

Step3: Calculate momentum for part 10

The formula for momentum is $p = mv$, where $m$ is the mass and $v$ is the velocity. Given $m = 8kg$ and $v = 25m/s$. Then $p=8\times25 = 200kg\cdot m/s$.

Step4: Calculate total momentum for part 11

First, find the total mass $M=m_{cyclist}+m_{bicycle}=43 + 7=50kg$. Then use the momentum formula $p = Mv$. Given $v = 5m/s$, so $p=50\times5=250kg\cdot m/s$.

Answer:

a) 190N to the right
b) $19m/s^{2}$

1. 200 kg(m)/s
2. 250 kg(m)/s