Homework Question: An 8.0 kg toy is dropped from a height of 7.0 m. What is the kinetic energy of the toy just before it hits the ground?
Well, our toy has an energy E, which is the sum of the potential and kinetic energy.So, when the toy is at the top, 7 m, it has potential energy, because it is at rest (its speed is 0)Ep = m*g*hWhere m is the mass (8 kg), g is the Earth's gravity (9.81 N/kg) and h is the height (7m). Please note that sometimes teachers use g = 10, in order to simplify the calculations.For the sake of simplicity, I will use 10. If you want to be exact, use 9.81.So, plugging in our numbers, we get:Ep = 8 kg * 7 m * 10 N/kg = 560 JWhen the toy falls, its energy is transforming from potential to kinetic.So, when the height is 0, all the energy is kinetic.Ek = 1/2 * m * v^2So, we know that energy is conserved, which means that the initial energy is the same as the final energy. We had calculated the initial energy to be 560 J, which means that Ek is also 560JSubstituting, we get that:1/2 * 8 kg * v^2 = 560V^2 = 140So, v is the square root of 140, which is 11.83 m/sFor any questions, feel free to comment!
A boy of mass 40 kg is standing on a weighing machine fixed to the floor of elevator. When the elevator begins to move up with constant acceleration a, the machine shows weight of the boy to be equal to 800 N. What's the value of a?
Based on Newton's second law. "Net force on a body equals product of its mass and acceleration." We need to find the net force I.e the resultant of all forces on the boy.We have been given the following forces1. Gravitational force downward =40g ~400N2. Normal reaction upward on the boy from the weighing machine=weight of the boy shown on the weighing machine=800NNet upward force on the boy=total external upward force-total external downward force=800N-400N=400NNet upward acceleration(again from second law)= 400/40= 10 m/s^2
A man of mass 60 kg climbs up a 20 m long staircase to the top of a building that’s 10 m high. What is the work done by him (g=10)?
Q: A man of mass 60 kg climbs up a 20 m long staircase to the top of a building that’s 10 m high. What is the work done by him (g=10)?Work is defined as:[math]W = Fd[/math]Work is also defined as:[math]W = F\Delta d[/math]Force is defined as:[math]F = ma[/math]Force is also defined as:[math]F = mg[/math]The integration of work is defined as:[math]W = \displaystyle \int_{10}^{20} F(x) \,dx[/math]The inverse of the integration of work (differentiation of force) is defined as:[math]F = \dfrac{dW}{dx}[/math]We can derive the formula defined as:[math]W = \left(mg \right)\Delta d[/math]Now, solve for the work of the object using the values given and using the formula derived above:[math]W = \left(60 \;kg \cdot 10 \;m/s^2 \right)\left(20 \;m-10 \;m \right)[/math][math]W = \left(600 \;kg \cdot m/s^2 \right)\left(10 \;m \right)[/math][math]W = 6000 \;J[/math][math]\therefore W = 6000 \;J[/math]
A block of mass 0.5 kg slides down a rough plane inclined at 30°, accelerating at 3ms^(-2). What's the friction between the block and the plane?
First draw a free body diagram of the block. A free body diagram shows all the forces acting on the object.Notice that I have defined a rotated set of axes and I labelled them x’ and y’. The x’-axis is parallel to the plane and the y’-axis is perpendicular to the plane. I chose positive x’-axis down the plane since the block is accelerating down the plane. Now write Newton’s second law in the x’ direction:[math]\Sigma F_{x'}=ma_{x'}[/math]The component of the weight (mg) acting down the plane is found by resolving the weight into components as shown below:So the component of the weight acting down the plane is (mg)sin30. The friction force acts opposite the direction of motion (up the plane) as shown on my free body diagram.[math]\Sigma F_{x'}=ma_{x'}[/math][math](mg)sin30-F_{fric}=ma_{x'}[/math][math](0.5)(9.81)sin30-F_{fric}=(0.5)(3\frac{m}{s^{2}})[/math][math]F_{fric}=0.953 N[/math]Once you know the friction force, you can determine the coefficient of friction using:[math]F_{fric}=\mu_{k} F_{N}[/math]To determine [math]F_{N}[/math] , write Newton’s second law in the y’ direction:[math]\Sigma F_{y'}=ma_{y'}[/math]Since the block is not “lifting” off the ramp, there is no motion in the y’-direction, so [math]a_{y'}=0.[/math][math]\Sigma F_{y'}=0[/math][math]F_{N}-(mg)cos30=0[/math][math]F_{N}=(0.5)(9.81)cos30 = 4.25 N[/math][math]F_{fric}=\mu_{k} F_{N}[/math][math]0.953N =\mu_{k} (4.25N)[/math][math]\mu_{k} = 0.224[/math]
How much force is needed to accelerate a 1000 kg car at a rate of 3m/s2?
If we use F=m×a or in this case F=1000×3So this car would need 3000 Newtons to accelerate it, discounting friction since we don't know what the surface is or the coefficient of friction.
An object has a mass of 5 kg. How much force is needed to accelerate it at 6 m/s2?
As per Newtons second law of motion,[math]F = m \times a[/math]Given data:Mass, [math]m = 5[/math] [math]kg[/math]Acceleration, [math]a = 6[/math] [math]m/s^2[/math]Therefore,[math]F = 5 \times 6[/math][math]F = 30[/math] NHence, [math]30[/math] [math]N[/math] of force is required to accelerate the object of [math]5[/math] [math]Kg[/math] to [math]6[/math] [math]m/s^2[/math]