Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the lawnmower is pushed using a massless handle, which makes an angle θ with the horizontal. Assume that Fh, the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the postive y direction to be upward. Use g for the magnitude of the acceleration due to gravity.

Part A

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Express the required force in terms of given quantities.

Fh =
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Part B

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Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

                                     F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

                                     F_net,x = 0

- The forces as follows:

                                     Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

                                     Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

                                    - W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

                                     Fn = W + Fh*sin(θ)

                                     Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

                                     Ff = u*Fn

                                     Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

                                     Fh*cos(θ) = u*(m*g + Fh*sin(θ))

                                     Fh*(cos(θ) - u*sin(θ)) = u*m*g

                                     Fh = u*m*g / (cos(θ) - u*sin(θ))

i would think it's food chain

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.