Understanding Force Problems: Comparing Forces on Driver and Passenger

[jcumby]

1.) A car moving at 50 km/h collides with a truck, and the front of the car is crushed 1.1 m as it comes to a complete stop. The driver is wearing a steabelt, but the pasenger is not. The passenger, obeying Newton's first law, keeps moving and slams into the dashboard after the car has stopped. If the dashboard compresses 5.0 cm on impact, find and compare the forces exerted on the driver by the seabelt and on the passenger by the dashboard. Assume the two have the same 65 kg mass.

I'm really confused about this one...thus far I have the intertia for both the driver and the passenger at 3250 N (65 kg * 50 km/h), but I'm not even sure if I've done that right.

2.)An elevator moves upward at 5.2 m/s. What is the minimum stopping time it can have if the passengers are to remain on the floor?

I know I need to figure out the normal force on a passenger on the verge of not remaining on the floor, and I know that normal force has to be some combination of mass and acceleration (9.8 + 5.2?), but other than that I'm stuck.

3.) Blocks of 1.0, 2.0, and 3.0 kg are lined up on a table (without friction). A 12 N force is applied to the 1.0 kg block. What force does the 2.0 kg block exert on the 3.0 kg block?

This must have something to do with inertia...but I'm not sure how to begin

 

[minifhncc]

1.) A car moving at 50 km/h collides with a truck, and the front of the car is crushed 1.1 m as it comes to a complete stop. The driver is wearing a steabelt, but the pasenger is not. The passenger, obeying Newton's first law, keeps moving and slams into the dashboard after the car has stopped. If the dashboard compresses 5.0 cm on impact, find and compare the forces exerted on the driver by the seabelt and on the passenger by the dashboard. Assume the two have the same 65 kg mass.

I'm really confused about this one...thus far I have the intertia for both the driver and the passenger at 3250 N (65 kg * 50 km/h), but I'm not even sure if I've done that right.

Convert units to SI and use equation v²=u²+2aS, after you find acceleration (deceleration in this case) for car (ie. also acceleration for driver, as driver moves with car body), and for the passenger, sub acceleration for driver and passenger into F=ma.

2.)An elevator moves upward at 5.2 m/s. What is the minimum stopping time it can have if the passengers are to remain on the floor?

I know I need to figure out the normal force on a passenger on the verge of not remaining on the floor, and I know that normal force has to be some combination of mass and acceleration (9.8 + 5.2?), but other than that I'm stuck.
[/quote]
Use equation a=(v-u)/t -- You know that a=9.8ms^-2, v=0 u= -5.2ms^-1, considering that down is positive.

3.) Blocks of 1.0, 2.0, and 3.0 kg are lined up on a table (without friction). A 12 N force is applied to the 1.0 kg block. What force does the 2.0 kg block exert on the 3.0 kg block?

This must have something to do with inertia...but I'm not sure how to begin

With no friction? Sounds like a dodgy question, anyone else agree?

Anyway, I'd start by finding out force due to gravity for each mass.

Hope this helps

 

[baba89]

.

I can provide some insights and guidance to help you understand and solve these force problems.

First, let's start with the car accident scenario. From what you have calculated, the initial inertia of both the driver and passenger is correct. However, to find the forces exerted on them, we need to use Newton's second law, which states that force is equal to mass multiplied by acceleration (F=ma). In this case, the acceleration is the change in velocity (from 50 km/h to 0 km/h) divided by the time it takes to stop. We can use the given distance and velocity to calculate the stopping time, and then plug that into the equation to find the forces.

For the driver, we know that the car stops in 1.1 m, so the stopping time can be calculated using the formula d=0.5at^2, where d is the distance, a is the acceleration, and t is the time. Rearranging this equation to solve for t, we get t=√(2d/a). Plugging in the values, we get t=√(2*1.1/50/3.6) = 0.059 seconds. Now, using F=ma, we can calculate the force exerted on the driver by the seatbelt: F=m*(Δv/t)=65*(50/3.6/0.059) = 5,934 N.

For the passenger, we can use the same process to calculate the stopping time, but we need to take into account the additional distance traveled (5.0 cm) before hitting the dashboard. This gives us a total distance of 1.15 m. Plugging this into the formula, we get t=√(2*1.15/50/3.6) = 0.061 seconds. Using F=ma again, we get F=m*(Δv/t)=65*(50/3.6/0.061) = 5,307 N.

Comparing these two forces, we can see that the force exerted on the passenger by the dashboard is slightly less than the force exerted on the driver by the seatbelt. This is due to the additional distance traveled by the passenger before coming to a complete stop.

Moving on to the elevator problem, we need to consider the normal force in order to determine the minimum stopping time. The normal force