A cat chases a mouse across a 1.1 m high
table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. The acceleration of gravity is 9.81m/s^2. What was the cat’s speed when it slid off the table? Answer in units of m/s.
To find the cat's speed when it slid off the table, we can use the principle of conservation of energy. The potential energy of the cat at the top of the table is converted into kinetic energy as it slides down.
Let's break down the problem into two parts: the cat's motion horizontally and vertically.
1. Vertical motion:
Since the mouse steps out of the way, the cat freely falls from the table height of 1.1 m. The potential energy at the top is converted to kinetic energy at the bottom.
The potential energy of the cat is given by the formula:
Potential energy = mass × gravity × height
Given that the height is 1.1 m and the acceleration due to gravity is 9.81 m/s^2, the potential energy becomes:
Potential energy = mass × 9.81 × 1.1
2. Horizontal motion:
The cat slides off the table and hits the floor 2.2 m from the edge. We can use the kinematic equation to find the speed of the cat.
The equation is:
Distance = initial velocity × time + (1/2) × acceleration × time^2
Since we need the speed when the cat slides off, we can assume the initial velocity is zero. Also, the time taken to fall can be calculated using the equation:
Distance = (1/2) × acceleration × time^2
Substituting the values, we have:
2.2 = (1/2) × 9.81 × time^2
Solving this equation gives us the time it takes for the cat to slide off the table.
Now, we can equate the potential energy from the vertical motion to the kinetic energy from the horizontal motion:
Potential energy = Kinetic energy
Since kinetic energy is given by the formula:
Kinetic energy = (1/2) × mass × velocity^2
Equating the potential energy to the kinetic energy, we get:
mass × 9.81 × 1.1 = (1/2) × mass × velocity^2
We can cancel out the mass, and solve for velocity:
9.81 × 1.1 = (1/2) × velocity^2
Solving this equation gives us the velocity of the cat when it slid off the table.
10c-<0.06>=f(n)(l1/l3+l2/l4)
uf=f(x)_ln<1.962>
fx/%fl
%fl=1.333
2C-<0.09>=f(1.33)
v1= __?