The figure below shows the forces exerted on an object that is sliding on a horizontal surface:Fg: the gravitational force of 40 N,Fn: the 40 N normal force exerted by the surface, andFf: a frictional force exerted to the left.If the coefficient of friction between the block and the surface is 0.15, then which of these most nearly represents the acceleration of the block? Group of answer choices2 m/s2 to the left1.5 m/s2 to the left2 m/s2 to the right1.5 m/s2 to the right
Question
The figure below shows the forces exerted on an object that is sliding on a horizontal surface:Fg: the gravitational force of 40 N,Fn: the 40 N normal force exerted by the surface, andFf: a frictional force exerted to the left.If the coefficient of friction between the block and the surface is 0.15, then which of these most nearly represents the acceleration of the block? Group of answer choices2 m/s2 to the left1.5 m/s2 to the left2 m/s2 to the right1.5 m/s2 to the right
Solution
First, we need to calculate the frictional force (Ff). The frictional force can be calculated using the formula:
Ff = μ * Fn
where μ is the coefficient of friction and Fn is the normal force. Given that μ is 0.15 and Fn is 40 N, we can substitute these values into the formula:
Ff = 0.15 * 40 = 6 N
The frictional force is acting to the left, opposing the motion of the object.
Next, we need to calculate the acceleration of the object using Newton's second law, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. The net force is the frictional force, and the mass can be calculated from the gravitational force (Fg = m * g, where g is the acceleration due to gravity, approximately 9.8 m/s²).
m = Fg / g = 40 N / 9.8 m/s² = 4.08 kg (approximately)
Now we can calculate the acceleration:
a = Ff / m = 6 N / 4.08 kg = 1.47 m/s²
So, the acceleration of the block is most nearly 1.5 m/s² to the left.
Similar Questions
A block of mass 6 kg is being pulled by force 24 N as shown. If coefficient of friction between block and the surface is 0.6, frictional force acting on the block is A.36 N B.32 NC.12 N D.none of these
.A block of weight 50N is moving over a horizontal surface starting at rest, moves (7)over a distance of 25m in l0 seconds under the action of a force of 20N. Determinethe coefficient of friction between the surfaces.18a䐀漀眀渀氀漀愀搀攀搀 昀爀漀洀 䬀琀甀渀漀琀攀猀⸀椀渀
Two blocks are arranged as below, connected together by rope and pulley with m1 = 3.50 kg and m2 = 7.00 kg. The coefficient of kinetic friction between all surfaces is 0.250. The top block is pulled to the right with a force F = 69.2 N. What is the acceleration of the top block?
A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6, If the acceleration of the truck is 5 m/s2, the frictional force acting on the block is :A 5 N B 6 N C 10 N D 15 N
A 2.2-kg block rests on a 30.0° incline as shown in the figure.The coefficient of static friction between the block and the incline is 0.63. There is a horizontal force force F of 6.4 N applied to the block. This is the maximum force possible without causing the block to slip. What is the normal force exerted on the block by the inclined surface?Express your answer in N, to at least one digit after the decimal point.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.