An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1 = 528 kg, m2 = 7.00 × 102 kg, and M3 = 311 kg have blocked a busy road. The rocks are lined up from left to right in the order m1, m2, and m3. The city calls a local contractor to use a bulldozer to clear the road. The bulldozer applies a constant force to m1 in order to slide the rocks off the road. Assuming the road is a flat, frictionless surface and the rocks are all in contact, what force, FA, must be applied to m¡ to slowly accelerate the group of rocks from the road at 0.100 m/s²? FA = N Use the value for FA to find the force, F12, exerted by m1 on m2. F12 N

The heaviest invertebrate is the giant squid, which is estimated to have a weight of about 0.36 tons spread out over its length of 37 feet. What is its weight in newtons? N

Tom and his little sister are enjoying an afternoon at the ice rink. They playfully place their hands together and push against each other. Tom's mass is 61 kg and his little sister's mass is 15 kg. (a) Which of the following statements is correct? The force experienced by the sister is less than the force experienced by Tom. They both experience the same force.     The force experienced by Tom is less than the force experienced by his sister. b) Which of the following statements is correct? Tom's acceleration is more than the sister's acceleration. They both have the same acceleration.     Tom's acceleration is less than the sister's acceleration.   c) If the sister's acceleration is 2.6 m/s2 in magnitude, what is the magnitude (in m/s2) of Tom's acceleration?

A frog with some sort of little monster on his back is swimming in a bathtub of split tomato soup. The frog's mass is m1 = 0.0280 kg, and the monster's mass is m2 = 0.0300 kg. The frog propels the monster and himself forward with a constant force of F = 0.440 N and accelerates at a constant rate of a = 2.65 m/s^2.  a) What is the resistive force of the split tomato soup on the frog? b) If the frog starts at rest at one side of the bathtub and swims across its length in t = 1.91 s, how long is the bathtub? c) How much work is done by the split tomato soup on the frog?

After falling from rest from a height of 29 m, a 0.45-kg ball rebounds upward, reaching a height of 19 m. If the contact between ball and ground lasted 2.2 ms, what average force was exerted on the ball? (Enter the magnitude.) N

Emma and Katie are racing along in their new wagon. Upon reaching the bottom of the hill traveling at 13.2 m/s, Katie notices her teddy bear 12.0 m in front of them, panics and locks the brakes sending the wagon into a skid. If the wagon and its contents have a total mass of 1.50 x10^1 kg, and the coefficient of kinetic friction between the road and the rubber wagon tires is (0.75), by how much does the wagon miss the bear?

A 54.9-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.93 m/s. Her hands are in contact with the wall for 1.09 s. Ignore friction and wind resistance. Find the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her). Note that this force has direction, which you should indicate with the sign of your answer. Number -9.72 Units

A student of mass 63 kg falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest at a time of 0.04 seconds. The average force exerted on him by the ground is 18,000 N, where the upward direction is taken to be a positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is only due to the ground

It may appear at first glance that Newton's 2nd and 3rd laws are independent claims. This, however, is not true - Newton's 3rd law is simply a consequence of Newton's 2nd law. In this problem, you will prove Newton's 3rd law for the normal contact force between two objects, using only Newton's 2nd law. Consider two blocks with masses m₁ and m₂ in contact, with external forces F₁ and F2 applied, as shown in the figure. The free body diagram for each block is also given, with internal forces fa and f, as shown. F₁ F₁ fa fb ΣF = ΣF = m₁ m2 Submit F₂ Complete the equations of motion for the entire system, as well as blocks 1 and 2 individually. Note, the forces have a signed magnitude, where positive values correspond to the positive x-direction ΣF= = (m₁ + m₂) a = m₁ a = m₂ a Use the above equations to write fa in terms of only fr: fa = m₁ You have used 0 of 4 attempts m₂ F₂ Save Calculator

A student (m = 58 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04 s. The average force exerted on him by the ground is +18000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. H=