Review. Two constant forces act on an object of mass m = 5.00 kg moving in the xy plane as shown in Figure P7.45. Force F, is 25.0 N at 35.0°, and force F, is 42.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.00î + 2.50j) m/s. (a) Express the two forces in unit-vector notation. Use unit-vector notation for your other answers. (b) Find the total force exerted on the object. (c) Find the object's acceleration. Now, considering the instant t = 3.00 s, find (d) velocity, (e) its position, (f) its from mv, and (g) its the object's kinetic energy F kinetic from 150° energy Jm Σ . Δr. (h) What conclusion can you draw by comparing the answers to parts (f) and (g)? 35.0° m Figure P7.45

Review. Two constant forces act on an object of mass m = 5.00 kg moving in the xy plane as shown in Figure P7.45. Force F, is 25.0 N at 35.0°, and force F, is 42.0 N at 150°. At time t = 0, the object is at the origin and has velocity (4.00î + 2.50j) m/s. (a) Express the two forces in unit-vector notation. Use unit-vector notation for your other answers. (b) Find the total force exerted on the object. (c) Find the object's acceleration. Now, considering the instant t = 3.00 s, find (d) velocity, (e) its position, (f) its from mv, and (g) its the object's kinetic energy F kinetic from 150° energy Jm Σ . Δr. (h) What conclusion can you draw by comparing the answers to parts (f) and (g)? 35.0° m Figure P7.45

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter8: Conservation Of Energy

Section: Chapter Questions

Problem 8.62AP: A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250...

See similar textbooks

Similar questions

A 1.00-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has a speed of vi = 3.00 m/s when it makes contact with a light spring (Fig. P8.62b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Fig. P8.62c). The object is then forced toward the left by the spring (Fig. P8.62d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Fig. P8.62e). Find (a) the distance of compression d, (b) the speed vat the unstretched posi-tion when the object is moving to the left (Fig. P8.624), and (c) the distance D where the abject comes to rest. Figure P8.62
A net force along the x-axis that has x-component Fx=−12.0N+(0.300N/m^2)x^2 is applied to a 4.60 kg object that is initially at the origin and moving in the −x direction with a speed of 8.60 m/s. A.what is the speed of the object when it reaches the point x=7.90m
A soccer ball with mass 0.460 kgkg is initially moving with speed 2.10 m/sm/s. A soccer player kicks the ball, exerting a constant force of magnitude 38.0 NN in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/sm/s? Express your answer in meters
At the time t1 = 0, an object with a mass m = 8.01 kg passes through the origin with a velocity v1 = (5.01î − 2.81ĵ) m/s. At a time t2 = 2.00 s, the object is at the position r = (8.00î + 4.00ĵ) m. Assume that the force acting on the object during the time interval Δt = t2 − t1 is constant and determine the following. (a) the kinetic energy of the object at the time t1 J(b) the force acting on the object during the time interval Δt = t2 − t1 (Express your answer in vector form.) F = N (c) the work done on the object by the force during the time interval Δt = t2 − t1 J(d) the kinetic energy of the object at the time t2 J(e) the speed of the object at the time t2 m/s
A net force along the x-axis that has x-component Fx = -12.0 N + (0.300 N/m2)x2 is applied to a 5.00 kg object that is initially at the origin and moving in the -x-direction with a speed of 6.00 m/s. What is the speed of the object when it reaches the point x = 5.00 m?
A net force along the x-axis that has x-component Fx = −12.0N + (0.300N/m2)x2 is applied to a 5.00 kg object that is initially at the origin and moving in the -x-direction with a speed of 6.00 m/s. What is the speed of the object when it reaches the point x = 5.00 m?
Mickey, a daredevil mouse of mass 0.0201 kg,0.0201 kg, is attempting to become the world's first "mouse cannonball." He is loaded into a spring‑powered gun pointing up at some angle and is shot into the air. The gun's spring has a force constant of 87.5 N/m87.5 N/m and is initially compressed a distance of 0.145 m0.145 m from its relaxed position. If Mickey has a constant horizontal speed of 2.15 m/s2.15 m/s while he is flying through the air, how high ℎh above his initial location in the gun does Mickey soar? Assume ?=9.81 m/s2.
An object of mass ?=8.9 kg is free to move on a horizontal surface without any friction. It is initially at rest and starts moving under a net force of ?⃗=(3 N)?̂+(4 N)?̂. What is the object's speed in m/s after moving 1 meter?
A net force along the x-axis that has x-component Fx=−12.0N+(0.300N/m^2)x^2 is applied to a 3.80 kg object that is initially at the origin and moving in the -x-direction with a speed of 7.30 m/s. What is the speed of the object when it reaches the point x = 8.20 m?
The only force acting on a 2.0-kg body as it moves along the x-axis is given by Fx = (12 - 2.0x) N, where x is in m. The velocity of the body at x = 2.0 m is 5.5 i m / s. What is the maximum kinetic energy that the body reaches? Answers: a) 36 J b) 39 J c) 43 J d) 46 J e) 30 J
At the time t1 = 0, an object with a mass m = 4.51 kg passes through the origin with a velocity v1 = (5.21î − 2.21ĵ) m/s. At a time t2 = 2.00 s, the object is at the position r = (8.00î + 4.00ĵ) m. Assume that the force acting on the object during the time interval Δt = t2 − t1 is constant and determine the following. the kinetic energy of the object at the time t1 the force acting on the object during the time interval Δt = t2 − t1(Express your answer in vector form.) the work done on the object by the force during the time interval Δt = t2 − t1 the kinetic energy of the object at the time t2 the speed of the object at the time t2
Luke Skywalker, 73 kg, is training on Dagobah. He runs up a incline for 10 m with an average velocity of 6 mph. He then performs the same run with Yoda on his back in 4.31 s. Assuming he makes the second run with equivalent power and that the acceleration due to gravity is the same as earth, what is Yoda's mass? 1 mi = 1.6 km A. 11 kg B. 13 kg C. 116 kg D. 189 kg