A box of mass 16 kg moves with a velocity which changes from (3, 0, –2) m/s to (3.01, –0.05, –1.96) m/s in a short time intervalAt = 0.01 s. Three forces are exerted on the box by objects in the surroundings, and these forces are constant during the short timeinterval. Two of these forces are F1 = (19, –13, –8) N and F 2 = (6,4, 2) N. What is the third force?É 3 = ( i)Nii

Answered: Image /qna-images/answer/2ee1c3f2-855e-42ad-b9cc-dc3e658aed4f.jpg

Answered: A box of mass 16 kg moves with a…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter5: Displacement And Force In Two Dimensions

Section: Chapter Questions

Problem 8STP

See similar textbooks

Similar questions

A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50×105 kg, its speed is 68.0 m/s, and the net braking force is 4.30×105N, what is its speed 11.0 s later? How far has it traveled in this time?
Three forces act on a 5-kg object: F1 = -2.0i + 3.0j N, F2 = 5.0i + 3.5j N, and F3 = 0.5i- 1.0j N. Assuming the object starts at rest at to = 0, write its velocity as a function of time.
A 3.5 kg particle moves along an x axis according to x(t) = −12 + 5t2 − 4t3, with xin meters and t in seconds. In unit-vector notation, what is the net force acting on theparticle at t = 6 s?
A box of mass 21 kg moves with a constant velocity of (4, 8, –4) m/s. Three forces are exerted on the box by objects in the surroundings. Two of these forces are F 1 = (1,0, 3) Nand É 2 = (-5, –1, 19) N. What is the third force? F3 = ( i ! )N i i
The block of mass m = 2 kg is released from the rest on the incline. If 0 = 15°, what is the velocity of the block 2 seconds after release? Ms = 0.3, k = 0.20 1.288 m/s 3.084 m/s 2.511 m/s m 0 Ө
A driver in a 1500-kg car driving at 10.0 m/s gently presses the brakes momentarily. The brakes apply a variable force, which can be modeled with the function F(t)=240tln(115t)F(t)=240tln⁡(115t), where F is in newtons and t is in seconds. What is the car's new speed after the brakes are held for 15.0 seconds, in m/s?
Two forces É, = (81.24i – 26.83j) N and É, = (27.41i - 80.52j) N are exerted on a particle. The particle's mass is 23.21 kg. (a) Find the particle's acceleration in component form. (Express your answer in vector form.) a = m/s? (b) What are the magnitude and direction of the acceleration? magnitude |m/s2 direction counterclockwise from the +x axis
A force F = (90, 80) N acts on a mass of 5 kg. At time t = 0 s, the mass is initially at rest. What are the (x,y) components of the velocity vector (in m/s) after 1.9 seconds?( , )
An constant applied force of 40.1 N lifts an 2.50 kg mass straight up, resulting in a final speed of 3.6 m/s. What distance did the mass move if it was initially at rest?
The 100 kg block in the figure takes 5.6 s to reach the floor after being released from rest. (Figure 1) Figure ✓ Part A m m = 100 kg 1 m What is the mass of the block on the left? Express your answer in kilograms. ΠΙ ΑΣΦ ? kg 1 of 1
A particle of mass 1.2 kg is acted on by forces F, = 18î N, F, = -11î N, and F, = 9.0j N. What is its acceleration (in m/s?) in this case? (Express your answer in vector form.) m/s?
A 1.4 gram bullet is shot into a tree stump. It enters at a velocity of 341 m/sec and comes to rest after having penetrated the stump in a straight line. It takes 0.00032 sec to come to a stop after striking the stump. What was the force on the bullet during impact? (answer in Newtons)

SEE MORE QUESTIONS

Recommended textbooks for you

Glencoe Physics: Principles and Problems, Student…

Physics

ISBN:

9780078807213

Author:

Paul W. Zitzewitz

Publisher:

Glencoe/McGraw-Hill