The strings and frictionless pulleys have negligible masses. The coefficients of kinetic friction is 0.25 and static friction is 0.30. Block A has a mass of 3 kg and block B has a mass of 4 kg. a) What is the maximum mass block C can have before block B will start to move? b) If block C has a mass of 7 kg and was 1 m above the ground, how long would it take block C to hit the ground starting from rest?

The strings and frictionless pulleys have negligible masses. The coefficients of kinetic friction is 0.25 and static friction is 0.30. Block A has a mass of 3 kg and block B has a mass of 4 kg. a) What is the maximum mass block C can have before block B will start to move? b) If block C has a mass of 7 kg and was 1 m above the ground, how long would it take block C to hit the ground starting from rest?

Similar questions

A car with a mass of 1500 kg is being towed by a rope held at a 20° angle to the horizontal. A friction force of 320 N opposes the car’s motion. What is the tension in the rope if the car goes from rest to 12 m/s in 10 s?
During a test crash, an air bag inflates to stop a dummy’s forward motion. The dummy’s mass is 75 kg. If the net force on the dummy is 825 N toward the rear of the car, what is the dummy’s deceleration?
At a post office, a parcel that is a 20.0-kg box slides down a ramp inclined at 28.3° with the horizontal. The coefficient of kinetic friction between the box and plane is 0.0300. a. Find the acceleration of the box. a = m/s? b. Find the velocity of the box as it reaches the end of the plane, if the length of the plane is 3 m and box starts at rest. Vf = m/s Submit Question
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.
You are in an elevator that is initially at rest. At t=0, it heads downward and speeds up as a function of time. It experiences a velocity given by v(t) = (−2.00t²ĵ)² m S a) What is the normal force that you experience at t=1.20 s? Assume you have a mass of 70.0 kg. b) How far did the elevator drop in 1.20 s?
A 5000 kg truck traveling 30 m/s comes to a stop. The coefficient of friction between the tires and the road is 0.80. What is the acceleration of the truck? What is the stopping distance?
A horizontal force, F1 = 55 N, and a force, F2 = 17 N acting at an angle of 0 to the horizontal, are applied to a block of mass m = 3.9 kg. The coefficient of kinetic friction between the block and the surface is uk 0.2. The block is moving to the right. = Randomized Variables F₁ = 55 N F₂=17 N m = 3.9 kg o vi ble. on d ed ed FN=46.72 (b) Solve numerically for the magnitude of acceleration of the block, a in m/s², if 0 = 30°. (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if 0 = 30°. ✓ Correct! a = 15.487 tan() П acos() E 14 sinh() 1 * 1 tanh() cotanh() Degrees O Radians 2 0 VO BACKSPACE sin() cos() cotan() asin() atan() acotan() cosh() Submit Hint . Feedback 7 85 9 6 3 I give up! • DEL HOME ← END y. CLEAR X Hk m 0
In the figure below, a box of Cheerios (mass mc = 1.5 kg) and a box of Wheaties (mass mw = 2.7 kg) are accelerated across a horizontal surface by a horizontal force F applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.0 N, and the magnitude of the frictional force on the Wheaties box is 4.1 N. If the magnitude of F is 12 N, what is the magnitude of the force on the Wheaties box from the Cheerios box? |---Select--- mc Mw
You drop an object of mass m from a tall building. Suppose the only forces affecting its motion are gravity, and air resistance proportional to the object's speed with positive constant of proportionality k. Let g denote gravitational acceleration (a positive constant). Express the total force in terms of m, g, and the object's velocity v, where upward displacement is considered positive. F = mg - kv Newton's second law tells us that force is equal to mass x acceleration, F = ma. Relating acceleration to velocity, rewrite the equation for total force above as a first order differential equation for v as a function of t. Denote v' as dv dt v(t) m(- dr) this is not an equation. Solve this differential equation for v(t) with the initial condition v(0) = V0. = mg (1-e ==) m k Find the terminal Terminal velocity = mg k X velocity. X syntax error: X X
A flea jumps by exerting a force of 1.17 x 10-5 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 1.12 x 10-6 N on the flea. Find the direction and magnitude (in m/s²) of the acceleration of the flea if its mass is 6.0 × 10-7 kg. (Let us assume that wind points to the right. We will consider this to be the +x direction and vertical to be the +y direction.) magnitude direction m/s² ° (measured clockwise from the vertical)
A 1500 kg car traveling 30 m/s comes to a stop. The coefficient of friction between the tires and the road is 0.80. What is the acceleration of the car? What is the stopping distance?
A student pushes a physics textbook of mass 1.15 kg across a table with a horizontal force of 5.48 Newtons. If the frictional force between the table and the textbook is 3.38 Newtons and the textbook was initially at rest, what is the speed of the textbook after 2.3 seconds? Remember to give your answer in metres per second and to at least 3 significant figures.