We can begin by resolving the gravitational force and the frictional force into their components parallel and perpendicular to the inclined plane:
- The gravitational force has a magnitude of mg, where m is the mass of the child and g is the acceleration due to gravity (9.81 m/s^2). The component parallel to the plane is mg*sin(42°), and the component perpendicular to the plane is mg*cos(42°).
- The frictional force has a magnitude of μN, where μ is the coefficient of kinetic friction and N is the normal force (the force perpendicular to the plane that prevents the child from sinking into it). Since the child is not sinking into the plane, the normal force has the same magnitude as the perpendicular component of the gravitational force, so N = mg*cos(42°).
The net force parallel to the plane is the difference between the parallel component of the gravitational force and the frictional force:
Fparallel = mg*sin(42°) - μN
Fparallel = mg*sin(42°) - μmg*cos(42°)
Fparallel = m(g*sin(42°) - μ*g*cos(42°))
The acceleration of the child down the slide is equal to the net force parallel to the plane divided by the mass of the child:
a = Fparallel / m
a = g*sin(42°) - μ*g*cos(42°)
a = 9.81*sin(42°) - 0.20*9.81*cos(42°)
a = 6.39 m/s^2
Therefore, the magnitude of the child's acceleration down the slide is 6.39 m/s^2.
At a playground a child slide down a slide that make a 42 degree angle with the horizontal direction.the coefficient of kinetic friction for this motion is 0.20 What is mangnitude of her accelation during her sliding
3 answers
A 5 kg block is pushed up a 30-degree incline plane with an applied force F parallel to the incline. calculate the amount of an applied force F needed to push a block up with acceleration of 2.0 m/s up the incline if the coefficient of friction is between surfaces is 0.30?
We can begin by drawing a free-body diagram of the forces acting on the block:
- The weight of the block (mg) acts downwards, perpendicular to the surface of the incline.
- The normal force (N) from the surface of the incline acts upwards, perpendicular to the surface of the incline.
- The force of friction (f) acts parallel to the incline, opposing the motion of the block.
- The applied force (F) acts parallel to the incline, in the same direction as the intended motion of the block.
Notice that we can resolve the weight of the block into components parallel and perpendicular to the incline using trigonometry. The weight perpendicular to the incline is m*g*cos(30°), and the weight parallel to the incline is m*g*sin(30°). Similarly, we can use the coefficient of friction to calculate the maximum force of friction the incline can exert on the block, which is μ*N. Since the block is moving up the incline, the direction of the frictional force is down the incline, opposite to the direction of intended motion.
We can write the equation for the net force acting on the block parallel to the incline:
F_parallel - f = ma
where a is the acceleration of the block up the incline. We can substitute the expressions for the forces parallel to the incline, and solve for F:
F - μ*N = ma
F - μ*m*g*cos(30°) = ma
F = ma + μ*m*g*cos(30°)
Now we need to substitute the given values and solve for F:
F = (5 kg)(2.0 m/s^2) + (0.30)(5 kg)(9.81 m/s^2)(cos 30°)
F = 10 N + 12.54 N
F = 22.54 N
Therefore, an applied force of at least 22.54 N is needed to push the block up the incline with an acceleration of 2.0 m/s^2, given the coefficient of friction between the surfaces is 0.30.
- The weight of the block (mg) acts downwards, perpendicular to the surface of the incline.
- The normal force (N) from the surface of the incline acts upwards, perpendicular to the surface of the incline.
- The force of friction (f) acts parallel to the incline, opposing the motion of the block.
- The applied force (F) acts parallel to the incline, in the same direction as the intended motion of the block.
Notice that we can resolve the weight of the block into components parallel and perpendicular to the incline using trigonometry. The weight perpendicular to the incline is m*g*cos(30°), and the weight parallel to the incline is m*g*sin(30°). Similarly, we can use the coefficient of friction to calculate the maximum force of friction the incline can exert on the block, which is μ*N. Since the block is moving up the incline, the direction of the frictional force is down the incline, opposite to the direction of intended motion.
We can write the equation for the net force acting on the block parallel to the incline:
F_parallel - f = ma
where a is the acceleration of the block up the incline. We can substitute the expressions for the forces parallel to the incline, and solve for F:
F - μ*N = ma
F - μ*m*g*cos(30°) = ma
F = ma + μ*m*g*cos(30°)
Now we need to substitute the given values and solve for F:
F = (5 kg)(2.0 m/s^2) + (0.30)(5 kg)(9.81 m/s^2)(cos 30°)
F = 10 N + 12.54 N
F = 22.54 N
Therefore, an applied force of at least 22.54 N is needed to push the block up the incline with an acceleration of 2.0 m/s^2, given the coefficient of friction between the surfaces is 0.30.